JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 5)
Let $$\overrightarrow{\mathrm{a}}=4 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=11 \hat{i}-\hat{j}+\hat{k}$$ and $$\overrightarrow{\mathrm{c}}$$ be a vector such that $$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}})$$.
If $$(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670$$, then $$|\vec{c}|^2$$ is equal to:
1600
1618
1627
1609
Explanation
$$\begin{aligned}
& \left.\begin{array}{l}
\vec{a}=4 \hat{i}-\hat{j}+\hat{k} \\
\vec{b}=11 \hat{i}-\hat{j}+\hat{k}
\end{array}\right] \\
& \vec{a} \cdot \vec{b}=44+1+1=46 \\
& |\vec{a}|^2=18,\left|\vec{b}^2\right|=123 \\
& (\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b}) \\
& (2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670 \\
& (\vec{a}+\vec{b}) \times c=(2 \vec{a}-3 \vec{b}) \times \vec{c} \\
& (-\vec{a}+4 \vec{b}) \times \vec{c}=0 \\
& \vec{c}=\lambda(4 \vec{b}-\vec{a}) \\
& (2 \vec{a}+3 \vec{b}) \cdot \lambda(4 \vec{b}-\vec{a})=1670 \\
& \lambda\left(5 \vec{a} \cdot \vec{b}-2|\vec{a}|^2+12|\vec{b}|^2\right)=1670 \\
& \lambda=\frac{1670}{5 \times 46-2 \times 18+12 \times 123} \\
& \lambda=1 \\
& \vec{c}=4 \vec{b}-\vec{a} \\
& =4(11 \hat{i}-\hat{j}+\hat{k})-(4 \hat{i}-\hat{j}+\hat{k}) \\
& =40 \hat{i}-3 \hat{j}+3 \hat{k} \\
& \left|\vec{c}^2\right|=1600+9+9 \\
& =1618 \\
\end{aligned}$$
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