Question

Let $\vec{\alpha } = \hat{i} + \hat{j} + \hat{k} , \vec{\beta} = \hat{i} - \hat{j} - \hat{k}$ and $\vec{\gamma} = - \hat{i} + \hat{j} - \hat{k}$ be three vectors. A vector $\vec{\delta}$, in the plane of $\vec{\alpha}$ and $\vec{\beta}$, whose projection on $\vec{\gamma}$ is $\frac{1}{\sqrt{3}}$, is given by

• A. $\hat{i} - 3 \hat{j} - 3 \hat{k}$
• B. $-\hat{3i} - 3 \hat{j} - \hat{k}$
• C. $\hat{3i} - \hat{j} + 3 \hat{k}$
• D. $\hat{i} + 3 \hat{j} - 3 \hat{k}$

Answer 1

Answer: (C)

$\hat{3i} - \hat{j} + 3 \hat{k}$

Answer 2

The correct answer is C:$3\hat{i}-\hat j+3\hat k$
Given that;
$\vec{a}=\hat i+\hat j+\hat k$
$\vec{b}=\hat i-\hat j+\hat k$
$\vec{c}=\hat i-\hat j-\hat k$
$\vec{v}$ on $\vec{a},\vec{b}$ whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$
So, Let $\vec{v}=\lambda \vec{a}+\mu \vec{b}$
$\vec{v}=(\lambda+\mu)\hat i+(\lambda-\mu)\hat j+(\lambda+\mu)\hat k$
projection of $\vec{v}$ on \vec{c}$$=\frac{\vec{v}.\vec{c}}{|\vec{c}|}=\frac{1}{\sqrt{3}}
⇒$\frac{(\lambda+\mu)-(\lambda-\mu)-(\lambda+\mu)}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
$\because(\vec{c}=\sqrt{1^2+(-1)^2+(-1)^2}=\sqrt{3})$
⇒$\mu-\lambda=1$
⇒$\vec{v}=(2\lambda+1)\hat{i}-\hat{j}+(2\lambda+1)\hat{k}$
For $\lambda=1$,$\vec{v}=3\hat{i}-\hat j+3\hat k$