Question

Which one of the following vectors is of magnitude $6$and perpendicular to both $a=4i-j+3k$ and $b=-2i+j-2k$?

• A. $2i-jk2k$
• B. $2\left(2i-j+2k\right)$
• C. $3\left(2i-j-2j\right)$
• D. $-2i+4j+4k$

Answer 1

The correct option is D

$-2i+4j+4k$

The explanation for the correct option

Option D:

Step 1: Find $a\times b$

The cross product of two vectors $A=a_{1}i+b_{1}j+c_{1}k$ and $B=a_{2}i+b_{2}j+c_{2}k$ is $A\times B=\left|\begin{array}{ccc}i& j& k\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$.

Given: $a=4i-j+3k$ and $b=-2i+j-2k$

To find the perpendicular vector, apply the cross product.

$\Rightarrow$$a\times b=\left|\begin{array}{ccc}i& j& k\\ 4& -1& 3\\ -2& 1& -2\end{array}\right|$

$\Rightarrow$$a\times b=i\left(2-3\right)-j\left(-8+6\right)+k\left(4-2\right)$

$\Rightarrow$$a\times b=-i+2j+2k$

Step 2: Find the magnitude of the vector

The magnitude of a vector $A=ai+bj+ck$ is $\sqrt{a^2+b^2+c^2}$

$\left|a\times b\right|=\sqrt{{\left(-1\right)}^2+{\left(2\right)}^2+{\left(2\right)}^2}$

$\Rightarrow$$\left|a\times b\right|=\sqrt{1+4+4}$

$\Rightarrow$$\left|a\times b\right|=3$

The required magnitude is $6$. So the vector is

$\frac{6}{3}\left(-i+2j+2k\right)$

$\Rightarrow$$2\left(-i+2j+2k\right)$

$\Rightarrow$$-2i+4j+4k$

Hence the correct option is D