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

The Correct Option is C

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 \(\lambda=1\),\(\vec{v}=3\hat{i}-\hat j+3\hat k\)