Question

If $\theta$ be the angle between the vectors $a=2i+2j-k$and$b=6i-3j+2k,$ then:

• A. $\cos \theta =\frac{4}{21}$
• B. $\cos \theta =\frac{3}{19}$
• C. $\cos \theta =\frac{2}{19}$
• D. $\cos \theta =\frac{5}{21}$

Answer 1

The correct option is A

$\cos \theta =\frac{4}{21}$

Explanation for the correct option:

Step 1. Find the value of $\left|a\right|$ and $\left|b\right|$.

For the vector $a=2i+2j-k$ the magnitude is given as:

$\begin{array}{rcl}\left|a\right|& =& \sqrt{2^2+2^2+{\left(-1\right)}^2}\\ & =& \sqrt{4+4+1}\\ & =& \sqrt{9}\\ & =& 3\end{array}$

For the vector $b=6i-3j+2k$ the magnitude is given as:

$\begin{array}{rcl}\left|b\right|& =& \sqrt{6^2+{\left(-3\right)}^2+2^2}\\ & =& \sqrt{36+9+4}\\ & =& \sqrt{49}\\ & =& 7\end{array}$

Step 2. Find the dot product.

For the vectors $a=2i+2j-k$and$b=6i-3j+2k$, the dot product $a·b$ is given as:

$\begin{array}{rcl}a·b& =& (2i+2j-k)·(6i-3j+2k)\\ & =& 2·6+2·(-3)+(-1)·2\\ & =& 12-6-2\\ & =& 4\end{array}$

Step 3. Find the value of $\cos \theta$.

The dot product is also given as: $a·b=\left|a\right|\left|b\right|\cos \theta$. So, $\cos \theta =\frac{a·b}{\left|a\right|\left|b\right|}$.

Substituting all the found values:

$\begin{array}{rcl}\cos \theta & =& \frac{a·b}{\left|a\right|\left|b\right|}\\ & =& \frac{4}{3\times 7}\left[a·b=4,\left|a\right|=3,\left|b\right|=7\right]\\ & =& \frac{4}{21}\end{array}$

Hence, the correct option is (A).