Let $\vec{a}=2 \hat{i}-3 \hat{j}+k, \vec{b}=3 \hat{i}+2 \hat{j}+5 k$ and a vector $\vec{c}$ be such that…

Let $\vec{a}=2 \hat{i}-3 \hat{j}+k, \vec{b}=3 \hat{i}+2 \hat{j}+5 k$ and a vector $\vec{c}$ be such that $(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=-18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \mathrm{k}$ and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3$. If $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{d}}$, then $|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}|$ is equal to :
  1. $18$
  2. $12$
  3. $9$
  4. $15$

Solution

$\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2 & -3 & 1 \\ 3 & 2 & 5\end{array}\right|$
$\begin{aligned} & =-17 \hat{i}-7 \hat{j}+13 \hat{k} \\ & (\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=-18 \hat{\mathrm{i}}-3 \mathrm{j}+12 \hat{k} \\ & \Rightarrow(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})-(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})=-18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}} \\ & \Rightarrow \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=(-18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{k})-(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \\ & =(-18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}})-(-17 \mathrm{i}-7 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & \therefore \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=(2 \mathrm{i}-3 \mathrm{j}+\mathrm{k}) \cdot(-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & =-2-12-1=-15 \\ & \therefore|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}|=15\end{aligned}$ ,

Asked in: JEE Main 2025 (02 Apr Shift 2)

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