JEE MAIN - Mathematics (2023 - 24th January Evening Shift - No. 17)
Let $$\overrightarrow a = \widehat i + 2\widehat j + \lambda \widehat k,\overrightarrow b = 3\widehat i - 5\widehat j - \lambda \widehat k,\overrightarrow a \,.\,\overrightarrow c = 7,2\overrightarrow b \,.\,\overrightarrow c + 43 = 0,\overrightarrow a \times \overrightarrow c = \overrightarrow b \times \overrightarrow c $$. Then $$\left| {\overrightarrow a \,.\,\overrightarrow b } \right|$$ is equal to :
Answer
8
Explanation
$$
\begin{aligned}
& \vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \vec{c}=7 \\\\
& \vec{a} \times \vec{c}-\vec{b} \times \vec{c}=\overrightarrow{0} \\\\
& (\vec{a}-\vec{b}) \times \vec{c}=0 \Rightarrow(\vec{a}-\vec{b}) \text { is paralleled to } \vec{c} \\\\
& \vec{a}-\vec{b}=\mu \vec{c}, \text { where } \mu \text { is a scalar } \\\\
& -2 \hat{i}+7 \hat{\mathrm{j}}+2 \lambda \hat{k}=\mu \cdot \overrightarrow{\mathrm{c}}
\end{aligned}
$$
Now $\vec{a} \cdot \overrightarrow{\mathbf{c}}=7$ gives $2 \lambda^2+12=7 \mu$
And $\vec{b} \cdot \vec{c}=-\frac{43}{2}$ gives $4 \lambda^2+82=43 \mu$
$\mu=2$ and $\lambda^2=1$
$$ |\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|=8 $$
Now $\vec{a} \cdot \overrightarrow{\mathbf{c}}=7$ gives $2 \lambda^2+12=7 \mu$
And $\vec{b} \cdot \vec{c}=-\frac{43}{2}$ gives $4 \lambda^2+82=43 \mu$
$\mu=2$ and $\lambda^2=1$
$$ |\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|=8 $$
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