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 -3 z =0\,\,\,\,\,\,\dots(i)$
$\vec{ v } \cdot(\hat{ i }+3 \hat{ j }-2 \hat{ k })=5$
$ \Rightarrow x +3 y -2 z =5\,\,\,\,\,\,\dots(ii)$
and $\vec{v} \cdot(2 \hat{i}+\hat{j}+4 \hat{k})=8$
$ \Rightarrow 2 x+y+4 z=8\,\,\,\,\,\,\dots(iv)$
Subtracting (ii) from (i), we have
$-2 y-z=-5$
$ \Rightarrow 2y+z=5\,\,\,\,\, \dots(iv)$
Multiply (ii) by 2 and subtracting (iii) from it, we obtain
$5 y-8 z=2\,\,\,\,\,\, \dots(v)$
Multiply (iv) by 8 and adding (v) to it, we have
$21 y=42$
$ \Rightarrow y=2\,\,\,\,\,\dots(v)$
Substituting $y =2 in ( iv )$, 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
$\vec{ v }= x \hat{ i }+\hat{ y }+ z \hat{ k }=\hat{ i }+2 \hat{ j }+\hat{ k }$