Question

Let vector $a=i-2j+k$ and vector $b=i-j+k$ be two vectors. If vector $c$ is a vector such that $bxc=bxa$ and $c·a=0$, then $c·b$ is equal to?

• A. $\frac{1}{2}$
• B. $-1$
• C. $-\frac{1}{2}$
• D. $-\frac{3}{2}$

Answer 1

The correct option is C

$-\frac{1}{2}$

Explanation for the correct answer:

Given, $a=i-2j+k$ and vector $b=i-j+k$

$$\begin{gathered} \left|a\right|=\sqrt{{\left(1\right)}^2+{(-2)}^2+{\left(1\right)}^2} \\ \left|a\right|=\sqrt{1+4+1} \\ \left|a\right|=\sqrt{6} \end{gathered}$$

$$\begin{gathered} \left|b\right|=\sqrt{{\left(1\right)}^2+{(-1)}^2+{\left(1\right)}^2} \\ \left|b\right|=\sqrt{1+1+1} \\ \left|b\right|=\sqrt{3} \end{gathered}$$

$$\begin{gathered} a.b=\left(1\right)\left(1\right)+(-2)(-1)+\left(1\right)\left(1\right) \\ a.b=1+2+1 \\ a.b=4 \end{gathered}$$

Now as per the given condition,

$$\begin{gathered} b\times c=b\times a \\ (b\times c)–(b\times a)=0 \\ b\times (c-a)=0 \end{gathered}$$

Thus, $b$ is parallel to $(c-a)$

$$\begin{gathered} c–a=kb \\ c=a+kb \end{gathered}$$

Also, given,

$$\begin{gathered} c·a=0 \\ (a+kb)a=0 \\ a·a+k(a·b)=0 \end{gathered}$$

so,

$$\begin{gathered} k=\frac{-|a{|}^2}{a·b} \\ k=-\frac{6}{4} \\ k=-\frac{3}{2} \end{gathered}$$

Therefore,

$$\begin{gathered} c=\frac{a-3}{2b} \\ c=(i-2j+k)-\frac{3}{2}(i-j+k) \\ c=\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k \\ c=-\frac{1}{2}(i+j+k) \end{gathered}$$

Hence,

$$\begin{gathered} b·c=-\frac{1}{2}(i+j+k)(i-j+k) \\ b·c=-\frac{1}{2} \end{gathered}$$

Hence, the correct option is (C)