Question

The magnitude of the projection of the vector $2\hat{i}+3\hat{j}+\hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$, is
A. $\dfrac{\sqrt{3}}{2}$
B. $\sqrt{\dfrac{3}{2}}$
C. $\sqrt{6}$
D. $3\sqrt{6}$

Answer 1

To solve the question, first of all we have to find the normal vector to the plane that contains the vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$. So, to find the normal vector to a plane having two vectors, we have to find the cross product of the two vectors. After finding the normal vector we have to find the projection of vector $2\hat{i}+3\hat{j}+\hat{k}$ on the normal vector using the projection equation, given by Projection of $\vec{a}$ on $\vec{b}$ is $=\left| \dfrac{~\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{b} \right|} \right|$ in the vector space.

Complete step by step answer:
It is given that there are two vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$ in the plane. So, first we find the normal vector of the plane containing these two vectors. We can find it by finding the cross product of $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$. That is,
$\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 1 & 1 & 1 \\\ 1 & 2 & 3 \\\ \end{matrix} \right|$
On solving, we get,
$\hat{i}\left( \left( 1\times 3 \right)-(1\times 2) \right)-\hat{j}\left( (1\times 3)-(1\times 1) \right)+\hat{k}\left( (1\times 2)-(1\times 1) \right)$
On expanding, we get,
$\hat{i}\left( 1 \right)-\hat{j}\left( 2 \right)+\hat{k}(1)$
$\Rightarrow \hat{i}-2\hat{j}+\hat{k}$
Therefore, the normal vector of the plane containing vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$ is $\hat{i}-2\hat{j}+\hat{k}$.
Now, we are asked to find the projection of vector $2\hat{i}+3\hat{j}+\hat{k}$ on the normal vector $\hat{i}-2\hat{j}+\hat{k}$. So, we can use the projection equation in the vector space. That is,
Projection of $\vec{a}$ on $\vec{b}$ is $=\left| \dfrac{~\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{b} \right|} \right|$
So, projection of $2\hat{i}+3\hat{j}+\hat{k}$ on $\hat{i}-2\hat{j}+\hat{k}$ = $\left| \dfrac{\left( 2\hat{i}+3\hat{j}+\hat{k} \right).\left( \hat{i}-2\hat{j}+\hat{k} \right)}{\sqrt{{{1}^{2}}+{{\left( -2 \right)}^{2}}+{{1}^{2}}}} \right|$
On solving we get,
Projection of $2\hat{i}+3\hat{j}+\hat{k}$ on $\hat{i}-2\hat{j}+\hat{k}$ = $\left| \dfrac{2-6+1}{\sqrt{1+4+1}} \right|$
Projection of $2\hat{i}+3\hat{j}+\hat{k}$ on $\hat{i}-2\hat{j}+\hat{k}$ $=\left| \dfrac{-3}{\sqrt{6}} \right|$
Therefore, the projection of $2\hat{i}+3\hat{j}+\hat{k}$ on $\hat{i}-2\hat{j}+\hat{k}$ $=\dfrac{3}{\sqrt{6}}$.
This can be further solved as $\dfrac{3}{\sqrt{6}}=\dfrac{\sqrt{3}\times \sqrt{3}}{\sqrt{3}\times \sqrt{2}}=\sqrt{\dfrac{3}{2}}$
So, the Projection of $2\hat{i}+3\hat{j}+\hat{k}$ on $\hat{i}-2\hat{j}+\hat{k}$ = $\sqrt{\dfrac{3}{2}}$.

So, the correct answer is “Option B”.

Note: We have to be careful while finding the normal vector because to find the normal vector of the plane we have to find the cross product of the vectors instead of the dot product. In the same way, we have to be careful while finding the projection because the equation for projection of $\vec{a}$ on $\vec{b}$ and projection of $\vec{b}$ on $\vec{a}$ are different. We have the projection of $\vec{b}$ on $\vec{a}$ as $=\left| \dfrac{~\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{a} \right|} \right|$ .