Q73 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics

a. $\vec{b}, \vec{c}$ are 3 vectors such that $|\vec{a}|=3,|\vec{b}|=2 \sqrt{2},|\vec{c}|=5$ and $\vec{c}$ is perpendicular to the plane of $\vec{a}$ and $\vec{b}$ If the angle between the vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$ then $|\vec{a}+\vec{b}+\vec{c}|=$

  • A
    $5 \sqrt{3}$
  • B
    $2 \sqrt{5}$
  • C
    10
  • Correct answer
    D
    $3 \sqrt{6}$

Solution

$|\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2 \vec{a} \cdot \vec{b}+2 \vec{c} \cdot \vec{a}+2 \vec{b} \cdot \vec{c}$ $\begin{aligned} & =9+8+25+2 \times|a| \cdot|b| \cos \left(\frac{\pi}{4}\right)+2 \times 0+2 \times 0 \\ & =9+8+25+2 \times 3 \times 2 \sqrt{2} \times \frac{1}{\sqrt{2}}=42+12=54 \\ & (\because \vec{a} \perp \vec{c} \Rightarrow \vec{a} \cdot \vec{c}=0 \text { and } \vec{b} \perp \vec{c} \Rightarrow \vec{b} \cdot \vec{c}=0) \\ & |\vec{a}+\vec{b}+\vec{c}|=\sqrt{54}=3 \sqrt{6}\end{aligned}$

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

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