In the previous section, we completed a discussion on dot product of two vectors. We saw projection of a vector also. In this section, we will see cross product of two vectors.
Vector (or cross) product of two vectors
We have already seen the basic details about vector product in chapter 7 of our physics classes. Let us recall:
1. We have seen right handed screws and left handed screws in section 7.10.
2. We have seen the direction of the cross product in section 7.11.
3. We have seen the magnitude of the cross product in section 7.12.
4. So we can write:
$\mathbf\small{\vec{a}\times \vec{b}~=~\left|\vec{a} \right|\left|\vec{b} \right|\sin\theta \,\hat{n}}$
Let us write the important properties of cross product:
1. $\mathbf\small{\vec{a}\times\vec{b}}$ is a vector
2. If $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ are two nonzero vectors, then $\mathbf\small{\vec{a}\times\vec{b}~=~\vec{0}}$ if and only if $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ are parallel (or collinear) to each other. We can write:
$\mathbf\small{\vec{a}\times\vec{b}~=~\vec{0}~\Leftrightarrow~\vec{a}{\parallel} \vec{b}}$
Let us see two specific cases:
Case 1:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}\times \vec{a}} & {~=~} &{\left|\vec{a} \right|\left|\vec{a} \right|\sin(0) \,\hat{n}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\left|\vec{a} \right|\left|\vec{a} \right|(0) \,\hat{n}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\vec{0}}
\\ \end{array}}$
Case 2:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}\times \left(-\vec{a} \right)} & {~=~} &{\left|\vec{a} \right|\left|-\vec{a} \right|\sin(\pi) \,\hat{n}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\left|\vec{a} \right|\left|-\vec{a} \right|(0) \,\hat{n}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\vec{0}}
\\ \end{array}}$
3. Let $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. Let the angle $\mathbf\small{\theta}$ between them be $\mathbf\small{\frac{\pi}{2}}$. Then we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}\times \vec{b}} & {~=~} &{\left|\vec{a} \right|\left|\vec{b} \right|\sin(\frac{\pi}{2}) \,\hat{n}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\left|\vec{a} \right|\left|\vec{b} \right|(1) \,\hat{n}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left|\vec{a} \right|\left|\vec{b} \right| \,\hat{n}}
\\ \end{array}}$
4. Based on (2) and (3), we get some interesting results:
• Based on (2), we get:
$\mathbf\small{\hat{i}\times\hat{i}~=~\hat{j}\times\hat{j}~=~\hat{k}\times\hat{k}~=~\vec{0}}$
• Based on (3), we get:
♦ $\mathbf\small{\hat{i}\times\hat{j}~=~\hat{k}}$
♦ $\mathbf\small{\hat{j}\times\hat{k}~=~\hat{i}}$
♦ $\mathbf\small{\hat{k}\times\hat{i}~=~\hat{j}}$
Also see fig.7.63 and fig.7.64(a) of section 7.12 of physics notes
5. Let $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. Let the angle $\mathbf\small{\theta}$ between them be $\mathbf\small{\theta}$. Then we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left|\vec{a}\times \vec{b} \right|} & {~=~} &{\left|\vec{a} \right|\left|\vec{b} \right|\sin\theta}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\sin \theta} & {~=~} &{\frac{\left|\vec{a}\times \vec{b} \right|}{\left|\vec{a} \right|\left|\vec{b} \right|}}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{ \theta} & {~=~} &{\sin^{-1}\left(\frac{\left|\vec{a}\times \vec{b} \right|}{\left|\vec{a} \right|\left|\vec{b} \right|} \right)}
\\ \end{array}}$
6. It is always true that, vector product is not commutative. This is because, $\mathbf\small{\vec{a}\times \vec{b}~=~-\left(\vec{b}\times \vec{a} \right)}$
• Assume that, both $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ lie on the plane of the paper.
• Also assume that, $\mathbf\small{\vec{a}\times\vec{b}}$ is directed towards the upper side of the plane of the paper.
• Then $\mathbf\small{\vec{b}\times\vec{a}}$ will be directed towards the bottom side of the plane of the paper.
• This is because:
♦ For $\mathbf\small{\vec{a}\times\vec{b}}$, we rotate the right handed screw from $\mathbf\small{\vec{a}~\text{to}~\vec{b}}$
♦ For $\mathbf\small{\vec{b}\times\vec{a}}$, we rotate the right handed screw from $\mathbf\small{\vec{b}~\text{to}~\vec{a}}$
7. Based on (4) and (6), we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\hat{i}\times \hat{j} } & {~=~} &{\hat{k}}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\hat{j}\times \hat{i}} & {~=~} &{-\left(\hat{i}\times \hat{j} \right)}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\hat{j}\times \hat{i}} & {~=~} &{-\hat{k}}
\\ \end{array}}$
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\hat{j}\times \hat{k} } & {~=~} &{\hat{i}}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\hat{k}\times \hat{j}} & {~=~} &{-\left(\hat{j}\times \hat{k} \right)}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\hat{k}\times \hat{j}} & {~=~} &{-\hat{i}}
\\ \end{array}}$
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\hat{k}\times \hat{i} } & {~=~} &{\hat{j}}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\hat{i}\times \hat{k}} & {~=~} &{-\left(\hat{k}\times \hat{i} \right)}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\hat{i}\times \hat{k}} & {~=~} &{-\hat{j}}
\\ \end{array}}$
8. The fig.26.30 below, shows triangle ABC.
 |
| Fig.26.30 |
• $\mathbf\small{\vec{AC}= \vec{a}~\text{and}~\vec{AB}=\vec{b}}$
• So $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ represent the adjacent sides of the triangle ABC
• We know that, area of the triangle ABC = $\mathbf\small{\frac{1}{2}\left(\left|\vec{AB} \right| \right)\left(CD \right)}$
• But $\mathbf\small{CD = \left|\vec{AC} \right|\sin\theta=\left|\vec{a} \right|\sin\theta}$
• So area of the triangle ABC
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{AB} \right| \right)\left(\left|\vec{a} \right|\sin\theta \right)}$
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{b} \right| \right)\left(\left|\vec{a} \right|\sin\theta \right)}$
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)}$
• We can write:
Area of the triangle is equal to half of the magnitude of the cross product
9. The fig.26.31 below, shows parallelogram ABCD.
 |
| Fig.26.31 |
• $\mathbf\small{\vec{AD}= \vec{a}~\text{and}~\vec{AB}=\vec{b}}$
• So $\mathbf\small{\vec{a}~\text{and}~\vec{b}}$ represent the adjacent sides of the parallelogram ABCD.
• We know that, area of the parallelogram ABCD = $\mathbf\small{\left(\left|\vec{AB} \right| \right)\left(DE \right)}$
• But $\mathbf\small{DE = \left|\vec{AD} \right|\sin\theta=\left|\vec{a} \right|\sin\theta}$
• So area of the parallelogram ABCD
= $\mathbf\small{\left(\left|\vec{AB} \right| \right)\left( \left|\vec{a} \right| \sin\theta \right)}$
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{b} \right| \right)\left( \left|\vec{a} \right|\sin\theta \right)}$
= $\mathbf\small{\left(\left|\vec{a}\times\vec{b} \right| \right)}$
• We can write:
Area of the parallelogram is equal to the magnitude of the cross product
10. Vector product of two vectors is also known as cross product of two vectors.
Two important properties of cross product
Property I: Distributivity of cross product over addition
This can be explained as below:
Let $\small{\vec{a},~\vec{b},~\vec{c}}$ be any three vectors. Then we can write:
$\small{\vec{a}\times\left(\vec{b}+\vec{c} \right)~=~\vec{a}\times\vec{b}+\vec{a}\times\vec{c}}$
Property II: Distributivity of cross product over multiplication
This can be explained as below:
Let $\small{\vec{a}~\text{and}~\vec{b}}$ be any two vectors and $\small{\lambda}$ be any scalar. Then we can write:
$\small{\lambda \left(\vec{a}\times\vec{b} \right) ~=~\left(\lambda \vec{a}\right)\times\vec{b} ~=~\vec{a}\times\left(\lambda\vec{b} \right)}$
Cross product when vectors are given in component form
• Let the two vectors be:
$\small{\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}}$
$\small{\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}}$
• Then we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}\times\vec{b}} & {~=~} &{\left(a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \right)\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{a_1\hat{i}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta} {} } &{} &{} & {} &{+~a_2\hat{j}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta} {} } &{} &{} & {} &{+~a_3\hat{k}\times\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \right)}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{a_1 b_1\left(\hat{i}\times\hat{i}\right)+a_1 b_2\left(\hat{i}\times\hat{j}\right)+a_1 b_3\left(\hat{i}\times\hat{k}\right)}
\\ {~\color{magenta} {} } &{} &{} & {} &{+~a_2 b_1\left(\hat{j}\times\hat{i}\right)+a_2 b_2\left(\hat{j}\times\hat{j}\right)+a_2 b_3\left(\hat{j}\times\hat{k}\right)}
\\ {~\color{magenta} {} } &{} &{} & {} &{+~a_3 b_1\left(\hat{k}\times\hat{i}\right)+a_3 b_2\left(\hat{k}\times\hat{j}\right)+a_3 b_3\left(\hat{k}\times\hat{k}\right)}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{a_1 b_1\left(\vec{0}\right)+a_1 b_2\left(\hat{k}\right)+a_1 b_3\left(-\hat{j}\right)}
\\ {~\color{magenta} {} } &{} &{} & {} &{+~a_2 b_1\left(-\hat{k}\right)+a_2 b_2\left(\vec{0}\right)+a_2 b_3\left(\hat{i}\right)}
\\ {~\color{magenta} {} } &{} &{} & {} &{+~a_3 b_1\left(\hat{j}\right)+a_3 b_2\left(-\hat{i}\right)+a_3 b_3\left(\vec{0}\right)}
\\ {~\color{magenta} 5 } &{} &{} & {~=~} &{\left(a_2 b_3 - a_3 b_2 \right)\hat{i} + \left(a_3 b_1 - a_1 b_3 \right)\hat{j} + \left(a_1 b_2 - a_2 b_1 \right)\hat{k}}
\\ {~\color{magenta} 6 } &{} &{} & {~=~} &{\left(a_2 b_3 - a_3 b_2 \right)\hat{i} - \left(a_1 b_3 - a_3 b_1 \right)\hat{j} + \left(a_1 b_2 - a_2 b_1 \right)\hat{k}}
\\ \end{array}}$
• The above result is equivalent to the determinant as shown below:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r} \hat{i} &{ \hat{j} } &{ \hat{k} } \\ a_{1} &{ a_{2} } &{ a_{3} }
\\ b_{1} &{ b_{2} } &{ b_{3} }
\\ \end{array}\right|$
Now we will see some solved examples.
Solved example 26.55
Find $\mathbf\small{\left|\vec{a}\times\vec{b} \right|}$, if $\mathbf\small{\vec{a}=2\hat{i}+\hat{j}+\hat{k}~\text{and}~\vec{b}=3\hat{i}+5\hat{j}-2\hat{k}}$
Solution:
1. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r} \hat{i} &{ \hat{j} } &{ \hat{k} } \\ 2 &{ 1 } &{ 3 }
\\ 3 &{ 5 } &{ -2 }
\\ \end{array}\right|$
2. So we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}\times\vec{b}} & {~=~} &{\left(-2 - 15 \right)\hat{i} - \left(-4 - 9 \right)\hat{j} + \left(10 - 3 \right)\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{-17\hat{i}+13\hat{j}+7\hat{k}}
\\ \end{array}}$
3. Therefore,
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left|\vec{a}\times\vec{b} \right|} & {~=~} &{\sqrt{(-17)^2 + (13)^2 + (7)^2}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\sqrt{507}}
\\ \end{array}}$
Solved example 26.56
Find
$\mathbf\small{\left|\vec{a}\times\vec{b} \right|}$, if
$\mathbf\small{\vec{a}=\hat{i}-7\hat{j}+7\hat{k}~\text{and}~\vec{b}=3\hat{i}-2\hat{j}+2\hat{k}}$
Solution:
1. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r} \hat{i} &{ \hat{j} } &{ \hat{k} } \\ 1 &{ -7 } &{ 7 }
\\ 3 &{ -2 } &{ 2 }
\\ \end{array}\right|$
2. So we get:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}}
&{\vec{a}\times\vec{b}} & {~=~} &{\left(-2 - 15
\right)\hat{i} - \left(-4 - 9 \right)\hat{j} + \left(10 - 3
\right)\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{-17\hat{i}+13\hat{j}+7\hat{k}}
\\ \end{array}}$
3. Therefore,
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left|\vec{a}\times\vec{b} \right|} & {~=~} &{\sqrt{(19)^2 + (19)^2}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{19\sqrt{2}}
\\ \end{array}}$
Solved example 26.57
Find a unit vector perpendicular to each of the vectors $\mathbf\small{\left(\vec{a}+\vec{b} \right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$, where $\mathbf\small{\vec{a}=\hat{i}+\hat{j}+\hat{k}~\text{and}~\vec{b}=\hat{i}+2\hat{j}+3\hat{k}}$
Solution:
1. We have:
• $\mathbf\small{\vec{c}=\left(\vec{a}+\vec{b} \right)=2\hat{i}+3\hat{j}+4\hat{k}}$
• $\mathbf\small{\vec{d}=\left(\vec{a}-\vec{b} \right)=-\hat{j}-2\hat{k}}$
2. We are asked to find a unit vector perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• We know that:
$\mathbf\small{\vec{c}\times\vec{d}}$ will be perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• So we will first write this cross product. We have:
$\vec{c}\times\vec{d}~=~\left|\begin{array}{r} \hat{i} &{ \hat{j} } &{ \hat{k} } \\ 2 &{ 3 } &{ 4 }
\\ 0 &{ -1 } &{ -2 }
\\ \end{array}\right|$
• So we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{c}\times\vec{d}} & {~=~} &{\left(-6 + 4 \right)\hat{i} - \left(-4 - 0 \right)\hat{j} + \left(-2 - 0 \right)\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{-2\hat{i}+4\hat{j}-2\hat{k}}
\\ \end{array}}$
3. Let $\mathbf\small{\vec{f}~=~\vec{c}\times\vec{d}}$
• Then $\mathbf\small{\hat{f}}$ is a required vector.
We have: $\mathbf\small{\hat{f}=\frac{\vec{f}}{\left|\vec{f} \right|}}$
= $\mathbf\small{\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{\sqrt{(-2)^2 + (4)^2+(-2)^2}}~=~\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{\sqrt{24}}}$
= $\mathbf\small{\frac{-2\hat{i}+4\hat{j}-2\hat{k}}{2\sqrt{6}}~=~\frac{-\hat{i}+2\hat{j}-\hat{k}}{\sqrt{6}}}$
4. Note:
We have a plane in which $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$ lie. The unit vector $\mathbf\small{\hat{f}}$ is perpendicular to that plane. Consequently, $\mathbf\small{-\hat{f}}$ will also be perpendicular to that plane. We would have obtained $\mathbf\small{-\hat{f}}$, if we had started with $\mathbf\small{\vec{d}\times\vec{c}}$ instead of $\mathbf\small{\vec{c}\times\vec{d}}$ in step (2)
Solved example 26.58
Find a unit vector
perpendicular to each of the vectors
$\mathbf\small{\left(\vec{a}+\vec{b}
\right)~\text{and}~\left(\vec{a}-\vec{b} \right)}$, where
$\mathbf\small{\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}~\text{and}~\vec{b}=\hat{i}+2\hat{j}-2\hat{k}}$
Solution:
1. We have:
• $\mathbf\small{\vec{c}=\left(\vec{a}+\vec{b} \right)=4\hat{i}+4\hat{j}}$
• $\mathbf\small{\vec{d}=\left(\vec{a}-\vec{b} \right)=-2\hat{i}+4\hat{k}}$
2. We are asked to find a unit vector perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• We know that:
$\mathbf\small{\vec{c}\times\vec{d}}$ will be perpendicular to both $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$
• So we will first write this cross product. We have:
$\vec{c}\times\vec{d}~=~\left|\begin{array}{r} \hat{i} &{ \hat{j} } &{ \hat{k} } \\ 4 &{ 4 } &{ 0 }
\\ -2 &{ 0 } &{ 4 }
\\ \end{array}\right|$
• So we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{c}\times\vec{d}} & {~=~} &{\left(16 - 0 \right)\hat{i} - \left(16 - 0 \right)\hat{j} + \left(0+8 \right)\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{16\hat{i}-16\hat{j}+8\hat{k}}
\\ \end{array}}$
3. Let $\mathbf\small{\vec{f}~=~\vec{c}\times\vec{d}}$
• Then $\mathbf\small{\hat{f}}$ is a required unit vector.
We have: $\mathbf\small{\hat{f}=\frac{\vec{f}}{\left|\vec{f} \right|}}$
= $\mathbf\small{\frac{16\hat{i}-16\hat{j}+8\hat{k}}{\sqrt{(16)^2 + (-16)^2+(8)^2}}~=~\frac{16\hat{i}-16\hat{j}+8\hat{k}}{24}}$
= $\mathbf\small{\frac{2\hat{i}-2\hat{j}+\hat{k}}{3}}$
4. Note:
We
have a plane in which $\mathbf\small{\vec{c}~\text{and}~\vec{d}}$ lie.
The unit vector $\mathbf\small{\hat{f}}$ is perpendicular to that plane.
Consequently, $\mathbf\small{-\hat{f}}$ will also be perpendicular to
that plane. We would have obtained $\mathbf\small{-\hat{f}}$, if we had
started with $\mathbf\small{\vec{d}\times\vec{c}}$ instead of
$\mathbf\small{\vec{c}\times\vec{d}}$ in step (2)
Solved example 26.59
Find the area of a triangle having the points A(1,1,1), B(1,2,3) and C(2,3,1) as it's vertices
Solution:
1. Fig.26.32 below shows the rough sketch:
 |
| Fig.26.32 |
2. Based on the rough sketch, we can write:
$\mathbf\small{\vec{a}= \vec{AB}=\hat{j}+2\hat{k}}$
$\mathbf\small{\vec{b}= \vec{AC}=\hat{i}+2\hat{j}}$
3. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r} \hat{i} &{ \hat{j} } &{ \hat{k} } \\ 0 &{ 1 } &{ 2 }
\\ 1 &{ 2 } &{ 0 }
\\ \end{array}\right|$
4. So we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}\times\vec{b}} & {~=~} &{\left(0 - 4 \right)\hat{i} - \left(0 - 2 \right)\hat{j} + \left(0 - 1 \right)\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{-4\hat{i}+2\hat{j}-\hat{k}}
\\ \end{array}}$
5. Then,
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left|\vec{a}\times\vec{b} \right|} & {~=~} &{\sqrt{(-4)^2 + (2)^2 + (-1)^2}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\sqrt{21}}
\\ \end{array}}$
6. Therefore,
Area of triangle ABC
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)~=~\frac{\sqrt{21}}{2}}$
Solved example 26.60
Find the area of a triangle with vertices A(1,1,2), B(2,3,5) and C(1,5,5).
Solution:
1. Fig.26.33 below shows the rough sketch:
 |
| Fig.26.33 |
2. Based on the rough sketch, we can write:
$\mathbf\small{\vec{a}= \vec{AB}=\hat{i}+2\hat{j}+3\hat{k}}$
$\mathbf\small{\vec{b}= \vec{AC}=4\hat{j}+3\hat{k}}$
3. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r} \hat{i} &{ \hat{j} } &{ \hat{k} } \\ 1 &{ 2 } &{ 3 }
\\ 0 &{ 4 } &{ 3 }
\\ \end{array}\right|$
4. So we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}\times\vec{b}} & {~=~} &{\left(6 - 12 \right)\hat{i} - \left(3 - 0 \right)\hat{j} + \left(4 - 0 \right)\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{-6\hat{i}-3\hat{j}+4\hat{k}}
\\ \end{array}}$
5. Then,
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left|\vec{a}\times\vec{b} \right|} & {~=~} &{\sqrt{(-6)^2 + (-3)^2 + (4)^2}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\sqrt{61}}
\\ \end{array}}$
6. Therefore,
Area of triangle ABC
= $\mathbf\small{\frac{1}{2}\left(\left|\vec{a}\times\vec{b} \right| \right)~=~\frac{\sqrt{61}}{2}}$
Solved example 26.61
Find the area of the parallelogram whose adjacent sides are given by the vectors $\small{\vec{a}=3\hat{i}+\hat{j}+4\hat{k}~\text{and}~\vec{b}=\hat{i}-\hat{j}+\hat{k}}$.
Solution:
1. Fig.26.34 below shows the rough sketch:
 |
| Fig.26.34 |
2. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r} \hat{i} &{ \hat{j} } &{ \hat{k} } \\ 3 &{ 1 } &{ 4 }
\\ 1 &{ -1 } &{ 1 }
\\ \end{array}\right|$
3. So we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}\times\vec{b}} & {~=~} &{\left(1 + 4 \right)\hat{i} - \left(3 - 4 \right)\hat{j} + \left(-3 - 1 \right)\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{5\hat{i}+\hat{j}-4\hat{k}}
\\ \end{array}}$
4. Then,
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left|\vec{a}\times\vec{b} \right|} & {~=~} &{\sqrt{(5)^2 + (1)^2 + (-4)^2}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\sqrt{42}}
\\ \end{array}}$
5. Therefore,
Area of parallelogram ABCD
= $\mathbf\small{\left|\vec{a}\times\vec{b} \right| ~=~\sqrt{42}}$
Solved example 26.62
Find the area of the
parallelogram whose adjacent sides are determined by the vectors
$\small{\vec{a}=\hat{i}-\hat{j}+3\hat{k}~\text{and}~\vec{b}=2\hat{i}-7\hat{j}+\hat{k}}$.
Solution:
1. Fig.26.35 below shows the rough sketch:
 |
| Fig.26.35 |
2. We have:
$\vec{a}\times\vec{b}~=~\left|\begin{array}{r} \hat{i} &{ \hat{j} } &{ \hat{k} } \\ 1 &{ -1 } &{ 3 }
\\ 2 &{ -7 } &{ 1 }
\\ \end{array}\right|$
3. So we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}\times\vec{b}} & {~=~} &{\left(-1 + 21 \right)\hat{i} - \left(1 - 6 \right)\hat{j} + \left(-7 + 2 \right)\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{20\hat{i}+5\hat{j}-5\hat{k}}
\\ \end{array}}$
4. Then,
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left|\vec{a}\times\vec{b} \right|} & {~=~} &{\sqrt{(20)^2 + (5)^2 + (-5)^2}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\sqrt{450}}
\\ \end{array}}$
5. Therefore,
Area of parallelogram ABCD
= $\mathbf\small{\left|\vec{a}\times\vec{b} \right| ~=~\sqrt{450}~=~15\sqrt{2}}$
Solved example 26.63
If a unit vector $\small{\vec{a}}$ makes angles $\small{\frac{\pi}{3}~\text{with}~\hat{i},~~\frac{\pi}{4}~\text{with}~\hat{j}}$ and an acute angle $\small{\theta}$ with $\small{\hat{k}}$, then find $\small{\theta}$ and hence, the components of $\small{\vec{a}}$.
Solution:
1. $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{3}~\text{with}~\hat{i}}$.
•
$\small{\hat{i}}$ lies along the x-axis. That means, $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{3}}$ with the x-axis.
•
So the x-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\frac{\pi}{3} \right)\hat{i}}$
= $\small{(1)\left(\frac{1}{2} \right)\hat{i}}$
= $\small{\left(\frac{1}{2} \right)\hat{i}}$
2. $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{4}~\text{with}~\hat{j}}$.
•
$\small{\hat{j}}$ lies along the y-axis. That means, $\small{\vec{a}}$ makes an angle $\small{\frac{\pi}{4}}$ with the y-axis.
•
So the y-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\frac{\pi}{4} \right)\hat{j}}$
= $\small{(1)\left(\frac{1}{\sqrt{2}} \right)\hat{j}}$
= $\small{\left(\frac{1}{\sqrt{2}} \right)\hat{j}}$
3. $\small{\vec{a}}$ makes an acute angle $\small{\theta~\text{with}~\hat{k}}$.
•
$\small{\hat{k}}$ lies along the z-axis. That means, $\small{\vec{a}}$ makes an acute angle $\small{\theta}$ with the z-axis.
•
So the z-component of $\small{\vec{a}}$
= $\small{\left|\vec{a} \right|\,\cos\left(\theta \right)\hat{k}}$
= $\small{(1)\cos\left(\theta \right)\hat{k}}$
= $\small{\cos\left(\theta \right)\hat{k}}$
4. We wrote the three components. So we can write:
$\small{\vec{a}=\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\cos\left(\theta \right)\hat{k}}$
5. Given that, $\small{\vec{a}}$ is a unit vector. So we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left|\vec{a} \right|} & {~=~} &{1}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\sqrt{\left(\frac{1}{2} \right)^2+\left(\frac{1}{\sqrt{2}} \right)^2+\left[\cos\left(\theta \right) \right]^2}} & {~=~} &{1}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\left(\frac{1}{2} \right)^2+\left(\frac{1}{\sqrt{2}} \right)^2+\left[\cos\left(\theta \right) \right]^2} & {~=~} &{1}
\\ {~\color{magenta} 4 } &{\Rightarrow} &{\cos^2\left(\theta \right)} & {~=~} &{1-\frac{1}{4}-\frac{1}{2}~=~\frac{1}{4}}
\\ {~\color{magenta} 5 } &{\Rightarrow} &{\cos\left(\theta \right)} & {~=~} &{\pm\frac{1}{2}}
\\ \end{array}}$
6. So we have two equations:
(i) $\small{\cos \theta~=~\frac{1}{2}}$
(ii) $\small{\cos \theta~=~-\frac{1}{2}}$
•
Solving the first equation, we get: $\small{\theta~=~\frac{\pi}{3}}$
•
Solving the second equation, we get: $\small{\theta~=~\frac{2\pi}{3}}$
•
Given that, $\small{\theta}$ is an acute angle. So we can write:
$\small{\theta~=~\frac{\pi}{3}}$
7. So based on step (4), we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}} & {~=~} &{\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\cos\left(\frac{\pi}{3} \right)\hat{k}}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\vec{a}} & {~=~} &{\left(\frac{1}{2} \right)\hat{i}+\left(\frac{1}{\sqrt{2}} \right)\hat{j}+\left(\frac{1}{2} \right)\hat{k}}
\\ \end{array}}$
8. Therefore, the components of $\small{\vec{a}}$ are:
$\small{\frac{1}{2},~\frac{1}{\sqrt{2}}~\text{and}~\frac{1}{2}}$
The link below gives a few more solved examples:
Exercise 10.4
In the next section, we will see some miscellaneous examples.
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