JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 9)
Let $$\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k}$$ and $$\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$$. Let $$\vec{d}$$ be a vector which is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$, and $$\vec{c} \cdot \vec{d}=12$$. Then $$(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$$ is equal to :
24
42
44
48
Explanation
If $\vec{d}$ is $\perp$ to both $\vec{a}$ and $\vec{b}$ then
$$ \vec{d}=\lambda(\vec{a} \times \vec{b})=\lambda\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 7 & -1 \\ 3 & 0 & 5 \end{array}\right|=(35 \hat{i}-13 \hat{j}-21 \hat{k}) \lambda $$
$$ \begin{aligned} & \text { but } \vec{c} \cdot \vec{d}=12 \Rightarrow \lambda(35 \times 1+13 \times 1-21 \times 2)=12 \\\\ & \Rightarrow \lambda(6)=12 \Rightarrow \lambda=2 \\\\ & \vec{\lambda}=2(35 \hat{i}-13 \hat{j}-21 \hat{k}) \end{aligned} $$
$$ \begin{aligned} & \text { Now, }(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d}) \\\\ & =\left|\begin{array}{ccc} -1 & 1 & -1 \\ 1 & -1 & 2 \\ 70 & -26 & -42 \end{array}\right| \\\\ & =-94+182-44=44 \end{aligned} $$
$$ \vec{d}=\lambda(\vec{a} \times \vec{b})=\lambda\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 7 & -1 \\ 3 & 0 & 5 \end{array}\right|=(35 \hat{i}-13 \hat{j}-21 \hat{k}) \lambda $$
$$ \begin{aligned} & \text { but } \vec{c} \cdot \vec{d}=12 \Rightarrow \lambda(35 \times 1+13 \times 1-21 \times 2)=12 \\\\ & \Rightarrow \lambda(6)=12 \Rightarrow \lambda=2 \\\\ & \vec{\lambda}=2(35 \hat{i}-13 \hat{j}-21 \hat{k}) \end{aligned} $$
$$ \begin{aligned} & \text { Now, }(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d}) \\\\ & =\left|\begin{array}{ccc} -1 & 1 & -1 \\ 1 & -1 & 2 \\ 70 & -26 & -42 \end{array}\right| \\\\ & =-94+182-44=44 \end{aligned} $$
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