Question

Vectors $\overrightarrow{a}=3\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-\hat{j}+2\hat{k}and\overrightarrow{c}=2\hat{i}-\hat{j}-\hat{k}$. Find vector $\overrightarrow{d}$if $\overline{d}$ is perpendicular to $\overrightarrow{c}$ and $\overrightarrow{d}$. $\overrightarrow{a}=10,\overrightarrow{b}=1$

Answer 1

$\overrightarrow{a}=3i+j+k$

$\overrightarrow{b}=i-j+2k$

$\overrightarrow{c}=2i-j-k$

$\overrightarrow{a}=10$

$\overrightarrow{b}=1$

Let $\overrightarrow{d}=xi+yj+zk$

Since $\overrightarrow{d}$ is perpendicular to $\overrightarrow{c}$.

Therefore,

$\overrightarrow{d}\cdot \overrightarrow{c}=0(\text{Dot product of}2\perp \text{vectors is}0)$

$\therefore (xi+yj+zk)\cdot (2i-j-k)=0$

$2x-y-z=0.....\left(1\right)$

$\overrightarrow{d}\cdot \overrightarrow{a}=10\left(\text{Given}\right)$

$\therefore (xi+yj+zk)\cdot (3i+j+k)=10$

$3x+y+z=10.....\left(2\right)$

$\overrightarrow{d}\cdot \overrightarrow{b}=1\left(\text{Given}\right)$

$\therefore (xi+yj+zk)\cdot (i-j+2k)=1$

$x-y+2z=1.....\left(3\right)$

Adding equation $\left(1\right)\&\left(2\right)$, we have

$(2x-y-z)+(3x+y+z)=0+10$

$5x=10$

$\Rightarrow x=2$

Now adding equation $\left(2\right)\&\left(3\right)$, we have

$(3x+y+z)+(x-y+2z)=10+1$

$4x+3z=11$

$4\left(2\right)+3z=11(\because x=2)$

$\Rightarrow z=\frac{11-8}{3}=1$

Substituting the value of $x$ and $z$ in equation $\left(1\right)$, we have

$2\left(2\right)-y-\left(1\right)=0$

$\Rightarrow y=3$

Therefore,

$\overrightarrow{d}=2i+3j+k$

Hence the value of $\overrightarrow{d}$ is $2i+3j+k$.