In the previous section, we saw components of a vector. We saw a solved example also. In this section, we will see a few more solved examples.
Solved example 26.7
Find the values of x and y so that the vectors $\small{2\hat{i} + 3\hat{j}}$ and $\small{x\hat{i} + y\hat{j}}$ are equal
Solution:
1. If two vectors are equal, their corresponding components will be equal.
2. Let us equate the corresponding components:
• Equating the vector components along the x-axis, we get: $\small{2\hat{i} = x\hat{i}}$. Therefore, x = 2
• Equating the vector components along the y-axis, we get: $\small{3\hat{j} = y\hat{i}}$. Therefore, y = 3
Solved example 26.8
Let $\small{\vec{a} = \hat{i} + 2\hat{j}}$ and $\small{\vec{b} = 2\hat{i} + \hat{j}}$.
Is $\small{\left|\vec{a} \right| = \left|\vec{b} \right|}$?
Are the vectors $\small{\vec{a}~\text{and}~\vec{b}}$ equal?
Solution:
Part (a): Comparing the magnitudes
1. $\small{\left|\vec{a} \right| = \sqrt{1^2 + 2^2} = \sqrt{5}}$
2. $\small{\left|\vec{b} \right| = \sqrt{2^2 + 1^2} = \sqrt{5}}$
3. The magnitudes are equal. So $\small{\left|\vec{a} \right| = \left|\vec{b} \right|}$
Part (b): Checking equality of vectors
1. If two vectors are equal, their corresponding components will be equal.
2. Let us compare the corresponding components:
• Comparing the vector components along the x-axis, we see that: $\small{\hat{i} ~\ne~ 2\hat{i}}$.
• Comparing the vector components along the y-axis, we see that: $\small{2\hat{j} ~\ne~ \hat{j}}$
3. Two vectors cannot be equal if even one of the corresponding components are not equal. So the given two vectors are not equal.
Solved example 26.9
Compute the magnitude of the following vectors:
$\small{\vec{a} = \hat{i} + \hat{j} + \hat{k}}$
$\small{\vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k}}$
$\small{\vec{c} = \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} - \frac{1}{\sqrt{3}} \hat{k}}$
Solution:
Part (a):
$\small{\left|\vec{a} \right| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}}$
Part (b):
$\small{\left|\vec{b} \right| = \sqrt{2^2 + (-7)^2 + (-3)^2} = \sqrt{4 + 49 + 9} = \sqrt{62}}$
Part (c):
$\small{\left|\vec{c} \right| = \sqrt{\left(\frac{1}{\sqrt{3}} \right)^2 + \left(\frac{1}{\sqrt{3}} \right)^2 + \left(-\frac{1}{\sqrt{3}} \right)^2} =\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1}$
Solved example 26.10
Write two different vectors having same magnitude
Solution:
We want two different vectors having the same magnitude. It can be done in 2 steps:
1. First write any convenient vector, say: $\small{\vec{a} = 3\hat{i}+4\hat{j}}$
2. Now we write the second vector by changing the order/sign of the scalar components. So the two vectors are:
$\small{\vec{a} = 3\hat{i}+4\hat{j}}$
$\small{\vec{b} = 4\hat{i}+3\hat{j}}$
• We have the magnitudes:
$\small{\left|\vec{a} \right| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5}$
$\small{\left|\vec{b} \right| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5}$
• Fig.26.20 below shows the difference in directions:
 |
| Fig.26.20 |
◼ Note:
There are other possibilities also:
$\small{\vec{a} = 3\hat{i}+4\hat{j}}$
$\small{\vec{c} = 3\hat{i}-4\hat{j}}$
$\small{\vec{d} = -4\hat{i}+3\hat{j}}$
$\small{\vec{e} = -3\hat{i}-4\hat{j}}$
• All these vectors have different directions. But they have the same magnitude. We can pick any two from them.
Solved example 26.11
Find the unit vector in the direction of the vector $\small{\vec{a} = 2\hat{i}+3\hat{j}+\hat{k}}$
Solution:
• We have: $\small{\hat{a}=\frac{\vec{a}}{\left|\vec{a} \right|}}$
• Thus we get:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\hat{a}} &
{~=~} &{\frac{2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{2^2 + 3^2 + 1^2}}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{14}}}
\\
{~\color{magenta} 3 } &{} &{} & {~=~}
&{\left(\frac{2}{\sqrt{14}} \right)
\hat{i}+\left(\frac{3}{\sqrt{14}} \right)
\hat{j}+\left(\frac{1}{\sqrt{14}} \right) \hat{k}}
\\ \end{array}}$
Solved example 26.12
Find the unit vector in the direction of the vector $\small{\vec{a} = \hat{i}+\hat{j}+2\hat{k}}$
Solution:
• We have: $\small{\hat{a}=\frac{\vec{a}}{\left|\vec{a} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\hat{a}} & {~=~} &{\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{1^2 + 1^2 + 2^2}}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left(\frac{1}{\sqrt{6}} \right) \hat{i}+\left(\frac{1}{\sqrt{6}} \right) \hat{j}+\left(\frac{2}{\sqrt{6}} \right) \hat{k}}
\\ \end{array}}$
Solved example 26.13
Find a vector in the direction of the vector $\small{\vec{a} = \hat{i}-2\hat{j}}$ that has magnitude 7 units.
Solution:
1. First we will write the unit vector $\small{\hat{a}}$
• We have: $\small{\hat{a}=\frac{\vec{a}}{\left|\vec{a} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\hat{a}} & {~=~} &{\frac{\hat{i}-2\hat{j}}{\sqrt{1^2 + (-2)^2}}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{\hat{i}-2\hat{j}}{\sqrt{5}}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left(\frac{1}{\sqrt{5}} \right) \hat{i}-\left(\frac{2}{\sqrt{5}} \right) \hat{j}}
\\ \end{array}}$
• This unit vector has the same direction as $\small{\vec{a}}$
2. So the required vector is:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{7\hat{a}} & {~=~} &{7\left[\left(\frac{1}{\sqrt{5}} \right) \hat{i}-\left(\frac{2}{\sqrt{5}} \right) \hat{j} \right]}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\left(\frac{7}{\sqrt{5}} \right) \hat{i}-\left(\frac{14}{\sqrt{5}} \right) \hat{j}}
\\ \end{array}}$
Solved example 26.14
Find a vector in the direction of the vector $\small{\vec{a} = 5\hat{i}-\hat{j}+2\hat{k}}$ that has magnitude 8 units.
Solution:
1. First we will write the unit vector $\small{\hat{a}}$
• We have: $\small{\hat{a}=\frac{\vec{a}}{\left|\vec{a} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\hat{a}} & {~=~} &{\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{5^2 + (-1)^2 + 2^2}}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{5\hat{i}-\hat{j}+2\hat{k}}{\sqrt{30}}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left(\frac{5}{\sqrt{30}} \right) \hat{i}-\left(\frac{1}{\sqrt{30}} \right) \hat{j}+\left(\frac{2}{\sqrt{30}} \right) \hat{j}}
\\ \end{array}}$
• This unit vector has the same direction as $\small{\vec{a}}$
2. So the required vector is:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{8\hat{a}} & {~=~} &{8\left[\left(\frac{5}{\sqrt{30}} \right) \hat{i}-\left(\frac{1}{\sqrt{30}} \right) \hat{j}+\left(\frac{2}{\sqrt{30}} \right) \hat{j} \right]}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\left(\frac{40}{\sqrt{30}} \right) \hat{i}-\left(\frac{8}{\sqrt{30}} \right) \hat{j}+\left(\frac{16}{\sqrt{30}} \right) \hat{j}}
\\ \end{array}}$
Solved example 26.15
Write two different vectors having same direction
Solution:
We want two different vectors having the same direction. It can be done in 6 steps:
1. First write any convenient vector, say: $\small{\vec{a} = 2\hat{i}+3\hat{j}}$
2. Now write $\small{\hat{a}}$. We get:
$\small{\hat{a}=\left(\frac{2}{\sqrt{13}} \right) \hat{i}+\left(\frac{3}{\sqrt{13}} \right) \hat{j}}$
• The reader may write all steps involved in finding the unit vector
3. Multiply the unit vector in (2), by any convenient scalar, to get a new vector $\small{\vec{b}}$:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{b}} & {~=~} &{\sqrt{13}\,\hat{a}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\sqrt{13}\left[\left(\frac{2}{\sqrt{13}} \right) \hat{i}+\left(\frac{3}{\sqrt{13}} \right) \hat{j} \right]}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\vec{b}} & {~=~} &{2 \hat{i}+3 \hat{j}}
\\ \end{array}}$
• Note that: $\small{\vec{b}=\vec{a}}$
4. Multiply the unit vector in (2), by any other convenient scalar, to get a third vector $\small{\vec{c}}$:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{c}} & {~=~} &{2\sqrt{13}\,\hat{a}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{2\sqrt{13}\left[\left(\frac{2}{\sqrt{13}} \right) \hat{i}+\left(\frac{3}{\sqrt{13}} \right) \hat{j} \right]}
\\ {~\color{magenta} 3 } &{\Rightarrow} &{\vec{c}} & {~=~} &{4 \hat{i}+6 \hat{j}}
\\ \end{array}}$
5. $\small{\vec{b}~\text{and}~\vec{c}}$ are the required vectors
6. Fig.26.21 shows the two vectors:
 |
| Fig.26.21 |
Solved example 26.16
Find the sum of the vectors
$\small{\vec{a} = \hat{i}-2 \hat{j} + \hat{k}}$
$\small{\vec{b} = -2\hat{i}+4 \hat{j} + 5\hat{k}}$
$\small{\vec{c} = \hat{i}-6 \hat{j} -7 \hat{k}}$
Solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{a}+\vec{b}+\vec{c}} & {~=~} &{(1-2+1)\hat{i}+(-2+4-6) \hat{j} + (1+5-7)\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{(0)\hat{i}+(-4) \hat{j} + (-1)\hat{k}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{-4 \hat{j}-\hat{k}}
\\ \end{array}}$
Solved example 26.17
Find unit vector in the direction of the sum of the vectors
$\small{\vec{a} = 2\hat{i}+2 \hat{j} -5 \hat{k}}$
$\small{\vec{b} = 2\hat{i}+ \hat{j} + 3\hat{k}}$
Solution:
1. First we will find the sum:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{c} = \vec{a}+\vec{b}} & {~=~} &{(2+2)\hat{i}+(2+1) \hat{j} + (-5+3)\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{(4)\hat{i}+(3) \hat{j} + (-2)\hat{k}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{4\hat{i}+3 \hat{j} -2\hat{k}}
\\ \end{array}}$
2. Now we will write the unit vector $\small{\hat{c}}$
• We have: $\small{\hat{c}=\frac{\vec{c}}{\left|\vec{c} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\hat{c}} & {~=~} &{\frac{4\hat{i}+3 \hat{j} -2\hat{k}}{\sqrt{4^2 + 3^2 + (-2)^2}}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{4\hat{i}+3 \hat{j} -2\hat{k}}{\sqrt{29}}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left(\frac{4}{\sqrt{29}} \right) \hat{i}+\left(\frac{3}{\sqrt{29}} \right) \hat{j}-\left(\frac{2}{\sqrt{29}} \right) \hat{k}}
\\ \end{array}}$
• This unit vector has the same direction as $\small{\vec{c}}$
Solved example 26.18
Find unit vector in the direction of the sum of the vectors
$\small{\vec{a} = 2\hat{i}- \hat{j} +2 \hat{k}}$
$\small{\vec{b} = -\hat{i}+ \hat{j} - \hat{k}}$
Solution:
1. First we will find the sum:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\vec{c} = \vec{a}+\vec{b}} & {~=~} &{(2-1)\hat{i}+(-1+1) \hat{j} + (2-1)\hat{k}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{(1)\hat{i}+(0) \hat{j} + (1)\hat{k}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\hat{i}+\hat{k}}
\\ \end{array}}$
2. Now we will write the unit vector $\small{\hat{c}}$
• We have: $\small{\hat{c}=\frac{\vec{c}}{\left|\vec{c} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\hat{c}} & {~=~} &{\frac{\hat{i}+\hat{k}}{\sqrt{1^2 + 1^2}}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{\hat{i}+\hat{k}}{\sqrt{2}}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left(\frac{1}{\sqrt{2}} \right) \hat{i}+\left(\frac{1}{\sqrt{2}} \right) \hat{k}}
\\ \end{array}}$
• This unit vector has the same direction as $\small{\vec{c}}$
Solved example 26.19
Write the direction ratios of the vector $\small{\vec{a} = \hat{i} + \hat{j} -2 \hat{k}}$ and hence calculate its direction cosines.
Solution:
1. Any vector is the resultant of three component vectors:
♦ Component along the OX axis, which is $\small{x \hat{i}}$
♦ Component along the OY axis, which is $\small{y \hat{j}}$
♦ Component along the OZ axis, which is $\small{z \hat{k}}$
• So any given vector can be written as: $\small{x\hat{i}+y\hat{j}+z\hat{k}}$
2. The vector given to us is: $\small{\vec{a}=\hat{i}+\hat{j}-2\hat{k}}$
• Comparing the corresponding components, we get: x = 1, y = 1 and z = −2
3. $\small{r = \left|\vec{a} \right|=\sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}}$
4. The three direction ratios are:
♦ $\small{lr = x = 1}$
♦ $\small{mr = y = 1}$
♦ $\small{nr = z = -2}$
Where $\small{l,~m,~n}$ are the direction cosines
5. So the three direction cosines are:
♦ $\small{l = \frac{x}{r} = \frac{1}{\sqrt{6}}}$
♦ $\small{m = \frac{y}{r} = \frac{1}{\sqrt{6}}}$
♦ $\small{n = \frac{z}{r} = \frac{-2}{\sqrt{6}}}$
Solved example 26.20
Find the direction cosines
of the vector $\small{\vec{a} = \hat{i} + 2\hat{j} +3 \hat{k}}$ and
hence calculate its direction cosines.
Solution:
1. Any vector is the resultant of three component vectors:
♦ Component along the OX axis, which is $\small{x \hat{i}}$
♦ Component along the OY axis, which is $\small{y \hat{j}}$
♦ Component along the OZ axis, which is $\small{z \hat{k}}$
• So any given vector can be written as: $\small{x\hat{i}+y\hat{j}+z\hat{k}}$
2. The vector given to us is: $\small{\vec{a}=\hat{i}+2\hat{j}+3\hat{k}}$
• Comparing the corresponding components, we get: x = 1, y = 2 and z = 3
3. $\small{r = \left|\vec{a} \right|=\sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}}$
4. The three direction ratios are:
♦ $\small{lr = x = 1}$
♦ $\small{mr = y = 2}$
♦ $\small{nr = z = 3}$
Where $\small{l,~m,~n}$ are the direction cosines
5. So the three direction cosines are:
♦ $\small{l = \frac{x}{r} = \frac{1}{\sqrt{14}}}$
♦ $\small{m = \frac{y}{r} = \frac{2}{\sqrt{14}}}$
♦ $\small{n = \frac{z}{r} = \frac{3}{\sqrt{14}}}$
Solved example 26.21
Show that the vector $\small{\vec{a} = \hat{i} + \hat{j} + \hat{k}}$ is equally inclined to the axes OX, OY and OZ.
Solution:
1. Any vector is the resultant of three component vectors:
♦ Component along the OX axis, which is $\small{x \hat{i}}$
♦ Component along the OY axis, which is $\small{y \hat{j}}$
♦ Component along the OZ axis, which is $\small{z \hat{k}}$
• So any given vector can be written as: $\small{x\hat{i}+y\hat{j}+z\hat{k}}$
2. The vector given to us is: $\small{\vec{a}=\hat{i}+\hat{j}+\hat{k}}$
• Comparing the corresponding components, we get: x = 1, y = 1 and z = 1
3. $\small{r = \left|\vec{a} \right|=\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}}$
4. The three direction ratios are:
♦ $\small{lr = x = 1}$
♦ $\small{mr = y = 1}$
♦ $\small{nr = z = 1}$
Where $\small{l,~m,~n}$ are the direction cosines
5. So the three direction cosines are:
♦ $\small{l = \frac{x}{r} = \frac{1}{\sqrt{3}}}$
♦ $\small{m = \frac{y}{r} = \frac{1}{\sqrt{3}}}$
♦ $\small{n = \frac{z}{r} = \frac{1}{\sqrt{3}}}$
6. The three direction cosines are equal. That means, the three angles are equal.
Solved example 26.22
Show that the vectors $\small{\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}}$ and
$\small{\vec{b}=-4\hat{i}+6\hat{j}-8\hat{k}}$ are collinear.
Solution:
1. Unit vector in the direction of $\small{\vec{a}}$
• We have: $\small{\hat{a}=\frac{\vec{a}}{\left|\vec{a} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\hat{a}} & {~=~} &{\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{2^2 + (-3)^2 + (4)^2}}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left(\frac{2}{\sqrt{29}} \right) \hat{i}-\left(\frac{3}{\sqrt{29}} \right) \hat{j}+\left(\frac{4}{\sqrt{29}} \right) \hat{k}}
\\ \end{array}}$
2. Unit vector in the direction of $\small{\vec{b}}$
• We have: $\small{\hat{b}=\frac{\vec{b}}{\left|\vec{b} \right|}}$
• Thus we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\hat{b}} & {~=~} &{\frac{-4\hat{i}+6\hat{j}-8\hat{k}}{\sqrt{(-4)^2 + 6^2 + (-8)^2}}}
\\ {~\color{magenta} 2 } &{} &{} & {~=~} &{\frac{-4\hat{i}+6\hat{j}-8\hat{k}}{\sqrt{116}}~=~\frac{-4\hat{i}+6\hat{j}-8\hat{k}}{2\sqrt{29}}}
\\ {~\color{magenta} 3 } &{} &{} & {~=~} &{\left(\frac{-2}{\sqrt{29}} \right) \hat{i}+\left(\frac{3}{\sqrt{29}} \right) \hat{j}-\left(\frac{4}{\sqrt{29}} \right) \hat{k}}
\\ \end{array}}$
3. We see that, $\small{\hat{a}=-\hat{b}}$
• That means: $\small{\hat{a}}$ has the exact opposite direction of $\small{\hat{b}}$
• That means: $\small{\vec{a}}$ has the exact opposite direction of $\small{\vec{b}}$
• That means: $\small{\vec{a}}$ and $\small{\vec{b}}$ are parallel.
• Therefore $\small{\vec{a}}$ and $\small{\vec{b}}$ are collinear.
Solved example 26.23
If $\small{\vec{a}~\text{and}~\vec{b}}$ are two collinear vectors, then which of the following are incorrect:
(a) $\small{\vec{b} = \lambda\vec{a}}$ for some scalar $\small{\lambda}$
(b) $\small{\vec{a} = \pm \vec{b}}$
(c) the respective components of $\small{\vec{a}~\text{and}~\vec{b}}$ are proportional
(d) both the vectors $\small{\vec{a}~\text{and}~\vec{b}}$ have same direction, but different magnitudes
Solution:
Part (a):
Given that $\small{\vec{a}~\text{and}~\vec{b}}$ are collinear. That means, they are parallel. So (a) is true.
Part (b):
$\small{\vec{a}~\text{and}~\vec{b}}$ are parallel. But they need not have the same direction. The directions may be opposite to each other. So (b) is true.
Part (c):
Since (a) is true, we can multiply each component of $\small{\vec{a}}$ by $\small{\lambda}$. That means, corresponding components are proportional. So (c) is true.
Part (d):
$\small{\vec{a}~\text{and}~\vec{b}}$ are parallel.
But they need not have the same direction. The directions may be opposite to
each other. So (d) is false.
Therefore, the correct option is (d)
In the next section, we will see vector joining two points.
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