Q50 / 75 · JEE Main 2025 (28 Jan Shift 1) · Mathematics
Let $\vec{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{d}}=\vec{a} \times \overrightarrow{\mathrm{b}}$. If $\overrightarrow{\mathrm{c}}$ is a vector such that $\vec{a} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|$, $|\overrightarrow{\mathrm{c}}-2 \vec{a}|^2=8$ and the angle between $\overrightarrow{\mathrm{d}}$ and $\overrightarrow{\mathrm{c}}$ is $\frac{\pi}{4}$, then $|10-3 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}|+|\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}|^2$ is equal to
Correct answer: 6
Solution
$\begin{aligned}
& \vec{a}=\hat{i}+\hat{j}+\hat{k} \\
& \vec{b}=2 \hat{i}+2 \hat{j}+\hat{k} \\
& \vec{d}=\vec{a} \times \vec{b} \\
& =-\hat{i}+\hat{j} \\
& |\vec{c}-2 \vec{a}|^2=8 \\
& |\vec{c}|^2+4|\vec{a}|^2-4 \vec{a} \cdot \vec{c}=8 \\
& |\vec{c}|^2+12-4|\vec{c}|=8 \\
& |\vec{c}|^2-4|\vec{c}|+4=0 \\
& |\vec{c}|^2=2 \\
& \vec{d}=\vec{a} \times \vec{b} \\
& \vec{d} \times \vec{c}=(\vec{a} \times \vec{b}) \times \vec{c} \\
& \left(|\vec{d}| \times|\vec{c}| \sin \frac{\pi}{4}\right)^2=((\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a})^2 \\
& 4=4|\vec{b}|^2+(\vec{b} \cdot \vec{c}) 2\left(|\vec{a}|^2\right)-2(\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{b})
\end{aligned}$
Let $\vec{b} \cdot \vec{c}=x$
$\begin{aligned}
& 4=36+3 x^2-20 x \\
& 3 x^2-20 x+32=0 \\
& x=\frac{8}{3}, 4 \\
& \Rightarrow \vec{b} \cdot \vec{c}=\frac{8}{3}, 4 \\
& \Rightarrow \vec{b} \cdot \vec{c}=\frac{8}{3}
\end{aligned}$
Now, $|10-3 \vec{b} \cdot \vec{c}|+|\vec{d} \times \vec{c}|^2$
$=|10-8|+(2)^2=6$
Hence, the correct answer is 6.