Question

The dot product of a vector with the vectors $\hat{i}+\hat{j}-3\hat{k}$, $\hat{i}+3\hat{j}-2\hat{k}$ and $2\hat{i}+\hat{j}-4\hat{k}$ are 0,5 and 8 respectively. Find the vector

• A. $\hat{i}+2\hat{j}+\hat{k}$
• B. $-\hat{i}+3\hat{j}-2\hat{k}$
• C. $\hat{i}+2\hat{j}+3\hat{k}$
• D. $\hat{i}-3\hat{j}-3\hat{k}$

Answer 1

Answer: (A)

Let the required vector be $\vec{v}=x\hat{i}+y\hat{j}+z\hat{k}$, then
$\vec{v}\cdot (\hat{i}+\hat{j}-3\hat{k})=0$
$\Rightarrow x+y-3z=0\dots (i)$
$\vec{v}\cdot (\hat{i}+3\hat{j}-2\hat{k})=5$
$\Rightarrow x+3y-2z=5\dots (ii)$
and $\vec{v}\cdot (2\hat{i}+\hat{j}+4\hat{k})=8$
$\Rightarrow 2x+y+4z=8\dots (iv)$
Subtracting (ii) from (i), we have
$-2y-z=-5$
$\Rightarrow 2y+z=5\dots (iv)$
Multiply (ii) by 2 and subtracting (iii) from it, we obtain
$5y-8z=2\dots (v)$
Multiply (iv) by 8 and adding (v) to it, we have
$21y=42$
$\Rightarrow y=2\dots (v)$
Substituting $y=2in(iv)$, we get
$2\times 2+z=5$
$\Rightarrow z=5-4=1$
Substituting these values in (i),we get
$x+2-3=0$
$\Rightarrow x=3-2=1$
Hence, the required vector is
$\vec{v}=x\hat{i}+\hat{y}+z\hat{k}=\hat{i}+2\hat{j}+\hat{k}$