# Let

Question:

Let $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$. Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$, and $\vec{c} \cdot \vec{d}=15$.

Solution:

Let $\vec{d}=d_{1} \hat{i}+d_{2} \hat{j}+d_{3} \hat{k}$

Since $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$, we have:

$\vec{d} \cdot \vec{a}=0$

$\Rightarrow d_{1}+4 d_{2}+2 d_{3}=0$            ...(1)

And,

$\vec{d} \cdot \vec{b}=0$

$\Rightarrow 3 d_{1}-2 d_{2}+7 d_{3}=0$        ...(2)

Also, it is given that:

$\vec{c} \cdot \vec{d}=15$

$\Rightarrow 2 d_{1}-d_{2}+4 d_{3}=15$            ...(3)

On solving (i), (ii), and (iii), we get:

$d_{1}=\frac{160}{3}, d_{2}=-\frac{5}{3}$ and $d_{3}=-\frac{70}{3}$

$\therefore \vec{d}=\frac{160}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{70}{3} \hat{k}=\frac{1}{3}(160 \hat{i}-5 \hat{j}-70 \hat{k})$

Hence, the required vector is $\frac{1}{3}(160 \hat{i}-5 \hat{j}-70 \hat{k})$.