JEE MAIN - Mathematics (2023 - 13th April Evening Shift - No. 14)
Let for a triangle $$\mathrm{ABC}$$,
$$\overrightarrow{\mathrm{AB}}=-2 \hat{i}+\hat{j}+3 \hat{k}$$
$$\overrightarrow{\mathrm{CB}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$$
$$\overrightarrow{\mathrm{CA}}=4 \hat{i}+3 \hat{j}+\delta \hat{k}$$
If $$\delta > 0$$ and the area of the triangle $$\mathrm{ABC}$$ is $$5 \sqrt{6}$$, then $$\overrightarrow{C B} \cdot \overrightarrow{C A}$$ is equal to
60
54
120
108
Explanation
_13th_April_Evening_Shift_en_14_1.png)
$$ \begin{aligned} & C A+A B=C B \\\\ & \Rightarrow +2 i+4 j+(\delta+3) k=\alpha i+\beta j+\gamma k \\\\ & \Rightarrow \alpha=+2, \beta=4, \gamma=\delta+3 \end{aligned} $$
$$ \begin{aligned} \text { Area } & =\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|=\left| {{1 \over 2}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 2} & 1 & 3 \cr { + 2} & 4 & \gamma \cr } } \right|} \right|=5 \sqrt{6} \\\\ & =(\gamma-12)^2+(6+2 \gamma)^2+100=(10 \sqrt{6})^2 \\\\ & \Rightarrow 5 \gamma^2=320 \end{aligned} $$
$$ \begin{aligned} \Rightarrow\gamma^2 =64 \\\\ \Rightarrow\gamma =8 \\\\ \Rightarrow\delta =5 \\\\ \overrightarrow{C B} \cdot \overrightarrow{C A} & =(2 i+4 j+8 k)(4 i+3 j+5 k) \\\\ & =8+12+40 \\\\ & =60 \end{aligned} $$
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