Let $a=i+j+k, b=i-j+k$ and $c=i-j-k$ be three vectors. A vector $v$ in the plane of a and $b$, whose projection on $c$ is $1 / \sqrt{3}$, is given by

Question

Let $a=i+j+k, b=i-j+k$ and $c=i-j-k$ be three vectors. A vector $v$ in the plane of a and $b$, whose projection on $c$ is $1 / \sqrt{3}$, is given by, $i-3 j-3 k$, $-3 i-3 j+k$, $3 i-j+3 k$, $i+3 j-3 k$

MARKS App · Accepted Answer

Given that, $$ \begin{array}{l} \vec{a}=\hat{i}+2 \hat{j}+\hat{k} \rightarrow(1) \ \vec{b}=\hat{i}-\hat{j}+\hat{k} \rightarrow(2) \ \vec{C}=\hat{i}+\hat{j}-\hat{k} \rightarrow(3) \end{array} $$ A vector in the plane of $ \vec{a} $ and $ \vec{b} $ can be written as, $$ \begin{array}{l} \vec{r}=m \vec{a}+n \vec{b} \ =m(\hat{i}+2 \hat{j}+\hat{k})+n(\hat{i}-\hat{j}+\hat{k}) \ \vec{r}=(m+n) \hat{i}+(2 m-n) \hat{j}+(m+n) \hat{k} \end{array} $$ So, projection on $ \vec{C} $ by $ \vec{r} $ is given by $$ \begin{array}{l} \frac{\vec{r} \cdot \vec{C}}{|\vec{C}|}=\frac{(m+n)+(2 m-n)-(m+n)}{\sqrt{1+1+1}} \ \Rightarrow \frac{2 m-n}{\sqrt{3}}=\frac{1}{\sqrt{3}} \end{array} $$ As use can see from Eq. (4), the $ ^{\wedge} i $ component and the $ \hat{k} $ component must be equal and $ \hat{j} $ can be either $ -1 $ or $ 1 . $ Therefore, option (4) is correct, that is $ 4 \hat{i}-\hat{j}+4 \hat{k} $

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