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JEE Mains 10 Apr 2023 Shift 2 Paper

Section: Mathematics

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Question : 18 of 90

Marks: +1, -0

Let

\vec{a}=2\hat{i}+7\hat{j}-\hat{k}, \vec{b}=3\hat{i}+5\hat{k}

\vec{C}=\hat{i}+\hat{j}+2\hat{k}

\vec{d}

-be a vector which is perpendicular to both

\vec{a}

\vec{b}

\vec{c}\cdot\vec{d}=12

(-\hat{i}+\hat{j}-\hat{k})\cdot(\vec{c}\times\vec{d})

[10-Apr-2023 shift 2]

24
42
48
44

Solution:

\;\vec{a}=2\hat{i}+7\hat{j}-\hat{k}

\;\vec{b}=3\hat{i}+5\hat{k}

\;\vec{c}=\hat{i}-\hat{j}+2\hat{k}

\;\vec{d}=\lambda(\vec{a}\times\vec{b})=\lambda\hat{\begin{vmatrix} i & \hat{j} & \hat{k} \\ 2 & 7 & -1 \\ 3 & 0 & 5 \end{vmatrix}}

\;\vec{d}=\lambda(35\hat{i}-13\hat{j}-21\hat{k})

\;\lambda(35+13-42)=12

\;\lambda=2

\;\vec{d}=2(35\hat{i}-13\hat{j}-21\hat{k})

\;(\hat{i}+\hat{j}-\hat{k})(\vec{c}\times\vec{d})

\;=\begin{vmatrix} -1 & 1 & -1 \\ 1 & -1 & 2 \\ 70 & -26 & -42 \end{vmatrix}=44

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