Question

Let $\overrightarrow{a}=4\hat{i}+5\hat{j}-\hat{k},\overrightarrow{b}=\hat{i}-4\hat{j}+5\hat{k}$ and $\overrightarrow{c}=3\hat{i}+\hat{j}-\hat{k}$. Find a vector $\overrightarrow{d}$ which is perpendicular to both $\overrightarrow{c}$ and $\overrightarrow{b}$ and $\overrightarrow{d}\cdot \overrightarrow{a}=21$.

Answer 1

Given:

$\overrightarrow{a}=4\hat{i}+5\hat{j}-\hat{k}$

$\overrightarrow{b}=\hat{i}-4\hat{j}+5\hat{k}$

$\overrightarrow{c}=3\hat{i}+\hat{j}-\hat{k}$

Let $\overrightarrow{d}=x\hat{i}+y\hat{j}+z\hat{k}$

Also given:

$\overrightarrow{d}.\overrightarrow{c}=0$ $\to ⟨1⟩$

$\overrightarrow{d}.\overrightarrow{b}=0$ $\to ⟨2⟩$ and

$\overrightarrow{d}.\overrightarrow{a}=21$ $\to ⟨3⟩$

Now, in $⟨1⟩$:

$(x\hat{i}+y\hat{j}+z\hat{k}).(3\hat{i}+\hat{j}-\hat{k})=0$

$\Rightarrow 3x+y-z=0$

Similarly from $⟨2⟩$ and $⟨3⟩:$

$x-4y+5z=0$ and

$4x+5y-z=21$

Solving these $e{q}^n$ in $x,y,z$ we get:

$x=-\frac{1}{3}$

$y=\frac{16}{3}$

$z=\frac{13}{3}$

Thus, the vector $\overrightarrow{d}=x\hat{i}+y\hat{j}+z\hat{k}$ can be written as :

$\overrightarrow{d}=-\frac{1}{3}\hat{i}+\frac{16}{3}\hat{j}+\frac{13}{3}\hat{k}$