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Let $\pi_1$ be the plane determined by the vectors $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}$ and $3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$. Let $\pi_2$ be the plane determined by the vectors $\hat{j}+2 \hat{k}$ and $3 \hat{k}-2 \hat{i}$. If $\theta$ is the angle between $\pi_1$ and $\pi_2$, then $\cos \theta=$
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Verified Answer
The correct answer is:
$-\frac{14}{29}$
Let $\vec{A}_1=\hat{i}+2 \hat{j}$ and $\vec{A}_2=3 \hat{j}-2 \hat{k}$ be the vectors related to $\pi_1$ and $\vec{B}_1=\hat{j}-2 \hat{k}, \vec{B}_2=3 \hat{k}-2 \hat{i}$ be the vectors related to $\pi_2$
Vector perpendicular to plane $\pi_1=\vec{n}_1=\vec{A}_1 \times \vec{A}_2$
$$
\vec{n}_1=-4 \hat{i}+2 \hat{j}+3 \hat{k}
$$
Vector perpendicular to plane $\pi_2=\vec{n}_2=\vec{B}_1 \times \vec{B}_2$
$$
\vec{n}_2=3 \hat{i}+4 \hat{j}+2 \hat{k}
$$
Hence
angle between $\pi_1$ and $\pi_2=$ angle between normals to the plane $\vec{n}_1$ and $\vec{n}_2$
$$
\begin{aligned}
& \therefore \cos \theta=\frac{\vec{n}_1 \cdot \vec{n}_2}{\left|\vec{n}_1\right|\left|\vec{n}_2\right|}=\frac{-12-8+6}{\sqrt{29} \sqrt{29}} \\
& \cos \theta=\frac{-14}{29}
\end{aligned}
$$
Vector perpendicular to plane $\pi_1=\vec{n}_1=\vec{A}_1 \times \vec{A}_2$
$$
\vec{n}_1=-4 \hat{i}+2 \hat{j}+3 \hat{k}
$$
Vector perpendicular to plane $\pi_2=\vec{n}_2=\vec{B}_1 \times \vec{B}_2$
$$
\vec{n}_2=3 \hat{i}+4 \hat{j}+2 \hat{k}
$$
Hence
angle between $\pi_1$ and $\pi_2=$ angle between normals to the plane $\vec{n}_1$ and $\vec{n}_2$
$$
\begin{aligned}
& \therefore \cos \theta=\frac{\vec{n}_1 \cdot \vec{n}_2}{\left|\vec{n}_1\right|\left|\vec{n}_2\right|}=\frac{-12-8+6}{\sqrt{29} \sqrt{29}} \\
& \cos \theta=\frac{-14}{29}
\end{aligned}
$$
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