In the previous section, we completed a discussion on the scalar product of two vectors. We saw some solved examples also. In this section, we will see a few more solved examples. Later in this section, we will see the Cauchy-Schwartz inequality.
Solved example 26.43
Evaluate the product $\small{\left(3\vec{a}-5\vec{b} \right).\left(2\vec{a}+7\vec{b} \right)}$.
Solution:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}}
&{\left(3\vec{a}-5\vec{b} \right).\left(2\vec{a}+7\vec{b}
\right)} & {~=~} &{3\vec{a}\left(2\vec{a}+7\vec{b}
\right)-5\vec{b}\left(2\vec{a}+7\vec{b} \right)}
\\
{~\color{magenta} 2 } &{} &{} & {~=~}
&{6\left|\vec{a}
\right|^2+21\vec{a}.\vec{b}-10\vec{b}.\vec{a}-35\left|\vec{b} \right|^2}
\\
{~\color{magenta} 3 } &{} &{} & {~=~}
&{6\left|\vec{a} \right|^2+11\vec{a}.\vec{b}-35\left|\vec{b}
\right|^2}
\\ \end{array}}$
Solved example 26.44
Find the magnitude of two vectors $\small{\vec{a}~\text{and}~\vec{b}}$,
having the same magnitude and such that the angle between them is
$\small{60^o}$ and their scalar product is $\small{\frac{1}{2}}$.
Solution:
1. Based on the given information, we can write:
♦ $\small{\left|\vec{a} \right|~=~\left|\vec{b} \right|}$
♦ $\small{\theta~=~60^o~=~\frac{\pi}{3}}$
♦ $\small{\vec{a}.\vec{b}~=~\frac{1}{2}}$
2. So we can write:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\vec{a}.\vec{b}}
& {~=~} &{\frac{1}{2}}
\\ {~\color{magenta} 2 }
&{\Rightarrow} &{\left|\vec{a} \right|\left|\vec{b}
\right|\cos\theta} & {~=~} &{\frac{1}{2}}
\\
{~\color{magenta} 3 } &{\Rightarrow} &{\left|\vec{a}
\right|^2 \cos\left(\frac{\pi}{3} \right)} & {~=~}
&{\frac{1}{2}}
\\ {~\color{magenta} 4 }
&{\Rightarrow} &{\left|\vec{a} \right|^2 \left(\frac{1}{2}
\right)} & {~=~} &{\frac{1}{2}}
\\ {~\color{magenta} 5 } &{\Rightarrow} &{\left|\vec{a} \right|^2} & {~=~} &{1}
\\
{~\color{magenta} 6 } &{\Rightarrow}
&{\left|\vec{a} \right|} & {~=~} &{1~=~\,\left|\vec{b}
\right|}
\\ \end{array}}$
◼ Remarks:
•
6 (magenta color): Here we discard the −ve root because, length cannot be −ve.
Solved example 26.45
Find $\small{\left|\vec{x} \right|}$ if for a unit vector
$\small{\vec{a}}$, $\small{\left(\vec{x}-\vec{a}
\right).\left(\vec{x}+\vec{a} \right)}$ = 12.
Solution:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\left(\vec{x}-\vec{a}
\right).\left(\vec{x}+\vec{a} \right)} & {~=~}
&{\vec{x}\left(\vec{x}+\vec{a} \right)-\vec{a}\left(\vec{x}+\vec{a}
\right)}
\\ {~\color{magenta} 2 } &{} &{}
& {~=~} &{\left|\vec{x}
\right|^2+\vec{x}.\vec{a}-\vec{a}.\vec{x}-\left|\vec{a} \right|^2}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left|\vec{x} \right|^2-\left|\vec{a} \right|^2}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{\left|\vec{x} \right|^2-(1)^2}
\\ {~\color{magenta} 5 } &{\Rightarrow} &{12} & {~=~} &{\left|\vec{x} \right|^2-1}
\\ {~\color{magenta} 6 } &{\Rightarrow} &{\left|\vec{x} \right|^2} & {~=~} &{13}
\\ {~\color{magenta} 7 } &{\Rightarrow} &{\left|\vec{x} \right|} & {~=~} &{\sqrt{13}}
\\ \end{array}}$
Solved example 26.46
If $\small{\vec{a}=2\hat{i}+2\hat{j}+3\hat{k}}$,
$\small{\vec{b}=-\hat{i}+2\hat{j}+\hat{k}}$ and
$\small{\vec{c}=3\hat{i}+\hat{j}}$ are such that
$\small{\left(\vec{a}+\lambda \vec{b} \right)}$ isperpendicular to
$\small{\vec{c}}$, then find the value of $\small{\lambda}$.
Solution:
1.First we write the sum:
$\small{\left(\vec{a}+\lambda \vec{b} \right)}$
= $\small{\vec{b}=\left(2-\lambda \right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(3+\lambda \right)\hat{k}}$
2. The resultant vector obtained above is perpendicular to $\small{\vec{c}}$. So their dot product will be zero. We can write:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\left[\left(2-\lambda
\right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(3+\lambda
\right)\hat{k} \right].\left[\vec{c} \right]} & {~=~} &{0}
\\
{~\color{magenta} 2 } &{\Rightarrow}
&{\left[\left(2-\lambda \right)\hat{i}+\left(2+2\lambda
\right)\hat{j}+\left(3+\lambda \right)\hat{k}
\right].\left[3\hat{i}+\hat{j} \right]} & {~=~} &{0}
\\
{~\color{magenta} 3 } &{\Rightarrow}
&{\left(2-\lambda \right)(3)+\left(2+2\lambda
\right)(1)+\left(3+\lambda \right)(0)} & {~=~} &{0}
\\ {~\color{magenta} 4 } &{\Rightarrow} &{6-3\lambda+2+2\lambda +0} & {~=~} &{0}
\\ {~\color{magenta} 5 } &{\Rightarrow} &{8-\lambda} & {~=~} &{0}
\\ {~\color{magenta} 6 } &{\Rightarrow} &{\lambda} & {~=~} &{8}
\\ \end{array}}$
Solved example 26.47
Show that $\small{\left|\vec{a} \right|\vec{b}+\left|\vec{b}
\right|\vec{a}}$,
is perpendicular to $\small{\left|\vec{a} \right|\vec{b}-\left|\vec{b}
\right|\vec{a}}$, for any two nonzero vectors
$\small{\vec{a}~\text{and}~\vec{b}}$.
Solution:
1. First we write the scalar product:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\left(\left|\vec{a}
\right|\vec{b}+\left|\vec{b} \right|\vec{a} \right).\left(\left|\vec{a}
\right|\vec{b}-\left|\vec{b} \right|\vec{a} \right)} & {~=~}
&{\left(\left|\vec{a} \right|\vec{b}+\left|\vec{b} \right|\vec{a}
\right).\left|\vec{a} \right|\vec{b}~-~\left(\left|\vec{a}
\right|\vec{b}+\left|\vec{b} \right|\vec{a} \right).\left|\vec{b}
\right|\vec{a}}
\\ {~\color{magenta} 2 } &{}
&{} & {~=~} &{\left|\vec{a} \right|^2\,\left|\vec{b}
\right|^2+\left(\left|\vec{a} \right|\left|\vec{b} \right|
\right)\vec{a}.\vec{b}-\left(\left|\vec{a} \right|\left|\vec{b} \right|
\right)\vec{b}.\vec{a}-\left|\vec{b} \right|^2\,\left|\vec{a} \right|^2}
\\
{~\color{magenta} 3 } &{} &{} & {~=~}
&{\left|\vec{a} \right|^2\,\left|\vec{b}
\right|^2+\left(\left|\vec{a} \right|\left|\vec{b} \right|
\right)\vec{a}.\vec{b}-\left(\left|\vec{a} \right|\left|\vec{b} \right|
\right)\vec{a}.\vec{b}-\left|\vec{b} \right|^2\,\left|\vec{a} \right|^2}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{0}
\\ \end{array}}$
2. Since the scalar product is zero, we can write:
The vectors are perpendicular to each other for any two nonzero vectors $\small{\vec{a}~\text{and}~\vec{b}}$
Solved example 26.48
If $\small{\vec{a}.\vec{a}=0}$ and $\small{\vec{a}.\vec{b}=0}$,
then what can be concluded about $\small{\vec{b}}$?
Solution:
1. We know that:
•
If $\small{\vec{p}.\vec{q}=0}$, then either
♦ $\small{\vec{p}}$ is a zero vector
♦ or $\small{\vec{q}}$ is a zero vector.
2. Given first information is that: $\small{\vec{a}.\vec{a}=0}$
•
It is clear that, $\small{\vec{a}}$ is a zero vector.
3. Now we consider the second information:
$\small{\vec{a}.\vec{b}=0}$
•
We found out that, $\small{\vec{a}}$ is zero vector. So $\small{\vec{b}}$ can be a zero vector or any nonzero vector.
Solved example 26.49
If $\small{\vec{a},~\vec{b},~\vec{c}}$ are unit vectors such that $\small{\vec{a}+\vec{b}+\vec{c}=\vec{0}}$,
find the value of $\small{\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}}$
Solution:
1. Given that: $\small{\vec{a}+\vec{b}+\vec{c}=\vec{0}}$
Multiplying both sides by $\small{\vec{a}}$, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}.\left[\vec{a}+\vec{b}+\vec{c} \right]} & {~=~} &{\vec{a}.\vec{0}}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{a}.\vec{c}} & {~=~} &{0}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{1+\vec{a}.\vec{b}+\vec{a}.\vec{c}} & {~=~} &{0}
\\ \end{array}}$
2. Multiplying both sides by $\small{\vec{b}}$, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{b}.\left[\vec{a}+\vec{b}+\vec{c} \right]} & {~=~} &{\vec{b}.\vec{0}}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\vec{b}.\vec{a}+\vec{b}.\vec{b}+\vec{b}.\vec{c}} & {~=~} &{0}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\vec{b}.\vec{a}+1+\vec{b}.\vec{c}} & {~=~} &{0}
\\ \end{array}}$
3. Multiplying both sides by $\small{\vec{c}}$, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{c}.\left[\vec{a}+\vec{b}+\vec{c} \right]} & {~=~} &{\vec{c}.\vec{0}}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+\vec{c}.\vec{c}} & {~=~} &{0}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+1} & {~=~} &{0}
\\ \end{array}}$
4. We have three results:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{1+\vec{a}.\vec{b}+\vec{a}.\vec{c}} & {~=~} &{0}
\\ {~\color{magenta} 2 } &{} &{\vec{b}.\vec{a}+1+\vec{b}.\vec{c}} & {~=~} &{0}
\\ {~\color{magenta} 3 } &{} &{\vec{c}.\vec{a}+\vec{c}.\vec{b}+1} & {~=~} &{0}
\\ \end{array}}$
5. Adding the L.H.S together, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{3 + 2\vec{a}.\vec{b} + 2\vec{b}.\vec{c} + 2\vec{c}.\vec{a}} & {~=~} &{0}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{3 + 2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right)} & {~=~} &{0}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{ 2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right)} & {~=~} &{-3}
\\ {~\color{magenta} 4 } &{\Rightarrow} &{ \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} } & {~=~} &{\frac{-3}{2}}
\\ \end{array}}$
Cauchy-Schwartz Inequality
This can be explained in 5 steps:
1. Let $\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. And let $\small{\theta}$ be the angle between them.
2. We can write the dot product and make some rearrangements as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}.\vec{b}} & {~=~} &{\left|\vec{a} \right|\left|\vec{b} \right|\cos\theta}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\frac{\vec{a}.\vec{b}}{\left|\vec{a} \right|\left|\vec{b} \right|}} & {~=~} &{\cos\theta}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\frac{\left|\left(\vec{a}.\vec{b} \right) \right|}{\left|\vec{a} \right|\left|\vec{b} \right|}} & {~=~} &{\left|\cos\theta \right|}
\\ \end{array}}$
◼ Remarks:
• 3 (magenta color): The denominator of the L.H.S is the product of two lengths. So the product will be +ve. We do not need to take the absolute value of that product
3. In the above result, the absolute value of $\small{\cos \theta}$ will be always less than or equal to 1. So we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{\left|\left(\vec{a}.\vec{b} \right) \right|}{\left|\vec{a} \right|\left|\vec{b} \right|}} & {~\le~} &{1}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\left|\left(\vec{a}.\vec{b} \right) \right|} & {~\le~} &{\left|\vec{a} \right|\left|\vec{b} \right|}
\\ \end{array}}$
4. So we can write:
♦ Absolute value of the dot product
♦ will be less than or equal to
♦ the product of the individual absolute values
• This is known as the Cauchy-Schwartz inequality
5. Note that, we used nonzero vectors to prove the triangle inequality. What if either $\small{\vec{a}~\text{or}~\vec{b}}$ is a zero vector?
The answer can be written in 2 steps:
(i) We have the inequality: $\small{\left|\vec{a}.\vec{b} \right| ~\le~\left|\vec{a} \right|\,\left|\vec{b} \right|}$
(ii) Suppose that, $\small{\vec{b}=\vec{0}}$. Then we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\left|\vec{a}.\vec{b} \right|} & {~\le~} &{\left|\vec{a} \right|\,\left|\vec{b} \right|}
\\ {~\color{magenta} 2 } &{\Rightarrow}&{\left|\vec{a}.\vec{0} \right|} & {~\le~} &{\left|\vec{a} \right|\,\left|\vec{0} \right|}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{0} & {~\le~} &{\left|\vec{a} \right|(0)}
\\ {~\color{magenta} 4 } &{\Rightarrow} &{0} & {~\le~} &{0}
\\ \end{array}}$
This is true.
Triangle Inequality
This can be explained in 12 steps:
1. Let $\small{\vec{a}~\text{and}~\vec{b}}$ be two nonzero vectors. And let their sum be $\small{\vec{c}}$.
2. We have:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{c}.\vec{c}} & {~=~} &{\left|\vec{c} \right|^2}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\left(\vec{a}+\vec{b} \right).\left(\vec{a}+\vec{b} \right)} & {~=~} &{\left|\left(\vec{a}+\vec{b} \right) \right|^2}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\left|\left(\vec{a}+\vec{b} \right) \right|^2} & {~=~} &{\left(\vec{a}+\vec{b} \right).\left(\vec{a}+\vec{b} \right)}
\\ {~\color{magenta} 4 } &{} &{} & {~=~} &{\vec{a}.\left(\vec{a}+\vec{b} \right)+\vec{b}.\left(\vec{a}+\vec{b} \right)}
\\ {~\color{magenta} 5 } &{} &{} & {~=~} &{\left|\vec{a} \right|^2+\vec{a}.\vec{b}+\vec{b}.\vec{a}+\left|\vec{b} \right|^2}
\\ {~\color{magenta} 6 } &{\Rightarrow} &{\left|\left(\vec{a}+\vec{b} \right) \right|^2} & {~=~} &{\left|\vec{a} \right|^2+2\vec{a}.\vec{b}+\left|\vec{b} \right|^2}
\\ \end{array}}$
3. Consider the middle term in the R.H.S of the above result.
• This term is $\small{2\vec{a}.\vec{b}}$. We know that, the dot product is a real number.
• Any real number will be less than or equal to it's absolute value. So we can write:
$\small{2\vec{a}.\vec{b}~\le~2\left|\vec{a}.\vec{b} \right|}$
4. Now compare $\small{\left[\left|\vec{a} \right|^2+2\vec{a}.\vec{b}+\left|\vec{b} \right|^2 \right]}$ and $\small{\left[\left|\vec{a} \right|^2+2\left|\vec{a}.\vec{b} \right|+\left|\vec{b} \right|^2 \right]}$
• Obviously, the former is less than or equal to the latter.
5. So we can proceed as follows:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left|\vec{a} \right|^2+2\vec{a}.\vec{b}+\left|\vec{b} \right|^2} & {~\le~} &{\left|\vec{a} \right|^2+2\left|\vec{a}.\vec{b} \right|+\left|\vec{b} \right|^2}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\left|\left(\vec{a}+\vec{b} \right) \right|^2} & {~\le~} &{\left|\vec{a} \right|^2+2\left|\vec{a}.\vec{b} \right|+\left|\vec{b} \right|^2}
\\ \end{array}}$
◼ Remarks:
• 2 (magenta color): Here we replace the L.H.S based on the result from (2)
6. Consider the middle term in the R.H.S of the above result.
• This term is $\small{2\left|\vec{a}.\vec{b} \right|}$.
• Based on Cauchy-Schwartz inequality, this term is less than or equal to $\small{2\left|\vec{a} \right|\left|\vec{b} \right|}$
7. Now compare $\small{\left[\left|\vec{a} \right|^2+2\left|\vec{a}.\vec{b} \right|+\left|\vec{b} \right|^2 \right]}$ and $\small{\left[\left|\vec{a} \right|^2+2\left|\vec{a} \right|\left|\vec{b} \right|+\left|\vec{b} \right|^2 \right]}$
• Obviously, the former is less than or equal to the latter.
8. So we can proceed as follows:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left|\left(\vec{a}+\vec{b} \right) \right|^2} & {~\le~} &{\left|\vec{a} \right|^2+2\left|\vec{a} \right|\left|\vec{b} \right|+\left|\vec{b} \right|^2}
\\ {~\color{magenta} 2 } &{\Rightarrow} &{\left|\left(\vec{a}+\vec{b} \right) \right|^2} & {~\le~} &{\left(\left|\vec{a} \right|+\left|\vec{b} \right| \right)^2}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\left|\left(\vec{a}+\vec{b} \right) \right|} & {~\le~} &{\left|\vec{a} \right|+\left|\vec{b} \right| }
\\ \end{array}}$
◼ Remarks:
• 2 (magenta color): Here we apply the identity:
(A+B)2 = A2 + 2AB + B2.
This identity is applicable because, every term in "magenta 1" is a real number.
9. So we can write:
♦ Magnitude of the "sum of two vectors"
♦ will be less than or equal to
♦ the sum of the individual magnitudes
• This is known as the triangle inequality
10. This inequality can be diagrammatically shown as in fig.26.28 below:
 |
| Fig.26.28 |
$\small{\vec{AC}}$ is the sum of $\small{\vec{AB}~\text{and}~\vec{BC}}$. Obviously, length of AC will be smaller than the sum of the lengths of AB and BC. We saw this in our earlier classes. Details here.
11. Consider the case when $\small{\left|\vec{a}+\vec{b} \right|=\left|\vec{a} \right|+\left|\vec{b} \right|}$
•
In such a situation, $\small{\left|\vec{AC} \right|=\left|\vec{AB} \right|+\left|\vec{BC} \right|}$
•
If the above equality is satisfied, the points A, B and C will be collinear.
12. Note that, we used nonzero vectors to prove the triangle inequality. What if either $\small{\vec{a}~\text{or}~\vec{b}}$ is a zero vector?
The answer can be written in 2 steps:
(i) We have the inequality: $\small{\left|\vec{a}+\vec{b} \right| ~\le~\left|\vec{a} \right|+\left|\vec{b} \right| }$
(ii) Suppose that, $\small{\vec{b}=\vec{0}}$. Then we can write:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\left|\left(\vec{a}+\vec{b} \right) \right|} & {~\le~} &{\left|\vec{a} \right|+\left|\vec{b} \right| }
\\ {~\color{magenta} 2 } &{\Rightarrow}&{\left|\left(\vec{a}+\vec{0} \right) \right|} & {~\le~} &{\left|\vec{a} \right|+\left|\vec{0} \right| }
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\left|\vec{a} \right|} & {~\le~} &{\left|\vec{a} \right|+0 }
\\ {~\color{magenta} 4 } &{\Rightarrow} &{\left|\vec{a} \right|} & {~\le~} &{\left|\vec{a} \right| }
\\ \end{array}}$
This is true.
Now we will see a solved example
Solved example 26.50
Show that the points
$\small{A\left(-2\hat{i}+3\hat{j}+5\hat{k} \right)}$
$\small{B\left(\hat{i}+2\hat{j}+3\hat{k} \right)}$
$\small{C\left(7\hat{i}-\hat{k} \right)}$
are collinear
Solution:
1. We are given the position vectors $\small{\vec{OA},~\vec{OB}~\text{and}~\vec{OC}}$
•
The three points form the vertices of a triangle ABC
2. Length of the side AB =
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\left|\vec{AB} \right|} & {=} &{\left|\vec{OB} ~-~\vec{OA}\right|}
\\ {~\color{magenta} 2 } &{}&{} & {=} &{\left|3\hat{i}-\hat{j}-2\hat{k} \right|}
\\ {~\color{magenta} 3 } &{}&{} & {=} &{\sqrt{14}}
\\ \end{array}}$
3. Length of the side BC =
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\left|\vec{BC} \right|} & {=} &{\left|\vec{OC} ~-~\vec{OB}\right|}
\\ {~\color{magenta} 2 } &{}&{} & {=} &{\left|6\hat{i}-2\hat{j}-4\hat{k} \right|}
\\ {~\color{magenta} 3 } &{}&{} & {=} &{\sqrt{56}~=~2\sqrt{14}}
\\ \end{array}}$
4. Length of the side AC =
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\left|\vec{AC} \right|} & {=} &{\left|\vec{OC} ~-~\vec{OA}\right|}
\\ {~\color{magenta} 2 } &{}&{} & {=} &{\left|9\hat{i}-3\hat{j}-6\hat{k} \right|}
\\ {~\color{magenta} 3 } &{}&{} & {=} &{\sqrt{126}~=~3\sqrt{14}}
\\ \end{array}}$
5. Now we analyze the above lengths. We see that:
$\small{\sqrt{14}~+~2\sqrt{14}~=~3\sqrt{14}}$
•
That means, $\small{\left|\vec{AB} \right|+\left|\vec{BC} \right|=\left|\vec{AC} \right|}$
•
So by triangle inequality, the points A, B and C are collinear
The above solved example helps us to obtain an interesting result. It can be explained in 4 steps:
1. First we write the sum:
$\small{\vec{AB}+\vec{BC}+\vec{CA}}$
= $\small{\left(3\hat{i}-\hat{j}-2\hat{k} \right)+\left(6\hat{i}-2\hat{j}-4\hat{k} \right)+(-1)\left(9\hat{i}-3\hat{j}-6\hat{k} \right)}$
= $\small{0\hat{i}+0\hat{j}+0\hat{k}}$
2. So we can write:
$\small{\vec{AB}+\vec{BC}+\vec{CA}~=~\vec{0}}$
3. Earlier, we saw that:
When sides of a triangle are taken in order, the resultant will be a null vector. See section 26.2
4. In our present case, resultant is a null vector. But the points do not form a triangle.
In the next section, we will see projection of a vector on a line.
Previous
Contents
Copyright©2026 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment