If a, b, c are three mutually perpendicular vectors such that the magnitudes of b and c are 1 / 2 times and √3 / 2 times that of a , respectively, then the angle between the vectors a+b+c and b is

Correct answer: (B) cos^-1 (1/(2 √2) ). Given, |b|=1/2|a| and |c|=(√3)/2|a| and a · b =b · c=a · c=0 |a+b+c|^2 =|a|^2+|b|^2+|c|^2=4|b|^2+|b|^2+3|b|^2=8|b|^2 ⇒|a+b+c| =2 √2|b| Angle between (a+b+c) and b is cos θ =((a+b+c) · b)/(|a+b+c||b|) ⇒ cos θ=(|b|^2)/(2 √2|b|^2)=1/(2 √2) θ =cos^-1 (1/(2 √2) )

Q72 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics

If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three mutually perpendicular vectors such that the magnitudes of $\mathbf{b}$ and $\mathbf{c}$ are $1 / 2$ times and $\sqrt{3} / 2$ times that of $\mathbf{a}$, respectively, then the angle between the vectors $\mathbf{a}+\mathbf{b}+\mathbf{c}$ and $\mathbf{b}$ is

A
$45^{\circ}$
Correct answer
B
$\cos ^{-1}\left(\frac{1}{2 \sqrt{2}}\right)$
C
$\cos ^{-1}\left(\frac{\sqrt{6}}{4}\right)$
D
$\cos ^{-1}\left(\frac{1}{4}\right)$

Solution

$\begin{aligned} & \text { Given, }|\mathbf{b}|=\frac{1}{2}|\mathbf{a}| \text { and }|\mathbf{c}|=\frac{\sqrt{3}}{2}|\mathbf{a}| \\ & \text { and } \quad \begin{aligned} \mathbf{a} \cdot \mathbf{b} & =\mathbf{b} \cdot \mathbf{c}=\mathbf{a} \cdot \mathbf{c}=0 \\ |\mathbf{a}+\mathbf{b}+\mathbf{c}|^2 & =|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2=4|\mathbf{b}|^2+|\mathbf{b}|^2+3|\mathbf{b}|^2=8|\mathbf{b}|^2 \\ \Rightarrow|\mathbf{a}+\mathbf{b}+\mathbf{c}| & =2 \sqrt{2}|\mathbf{b}| \end{aligned} \end{aligned}$ Angle between $(\mathbf{a}+\mathbf{b}+\mathbf{c})$ and $\mathbf{b}$ is $\begin{aligned} \cos \theta & =\frac{(\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot \mathbf{b}}{|\mathbf{a}+\mathbf{b}+\mathbf{c}||\mathbf{b}|} \Rightarrow \cos \theta=\frac{|\mathbf{b}|^2}{2 \sqrt{2}|\mathbf{b}|^2}=\frac{1}{2 \sqrt{2}} \\ \theta & =\cos ^{-1}\left(\frac{1}{2 \sqrt{2}}\right) \end{aligned}$

Hence, the correct answer is option: 'B'.

Solution

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