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(B) Let $\vec{A}_1 = \hat{i} + 2\hat{j}$ and $\vec{A}_2 = 3\hat{j} - 2\hat{k}$ be the vectors defining plane $\pi_1$. The normal vector $\vec{n}_1$ to

Vedclass · Accepted Answer

$-\frac{14}{29}$

Vedclass · Answer

(B) Let $\vec{A}_1 = \hat{i} + 2\hat{j}$ and $\vec{A}_2 = 3\hat{j} - 2\hat{k}$ be the vectors defining plane $\pi_1$. The normal vector $\vec{n}_1$ to $\pi_1$ is given by the cross product $\vec{A}_1 	imes \vec{A}_2$: $\vec{n}_1 = (\hat{i} + 2\hat{j}) 	imes (3\hat{j} - 2\hat{k}) = -4\hat{i} + 2\hat{j} + 3\hat{k}$. Let $\vec{B}_1 = \hat{j} + 2\hat{k}$ and $\vec{B}_2 = 3\hat{k} - 2\hat{i}$ be the vectors defining plane $\pi_2$. The normal vector $\vec{n}_2$ to $\pi_2$ is given by the cross product $\vec{B}_1 	imes \vec{B}_2$: $\vec{n}_2 = (\hat{j} + 2\hat{k}) 	imes (3\hat{k} - 2\hat{i}) = 3\hat{i} + 4\hat{j} + 2\hat{k}$. The angle $	heta$ between the planes $\pi_1$ and $\pi_2$ is the angle between their normal vectors $\vec{n}_1$ and $\vec{n}_2$. $\cos 	heta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$. $\vec{n}_1 \cdot \vec{n}_2 = (-4)(3) + (2)(4) + (3)(2) = -12 + 8 + 6 = 2$. $|\vec{n}_1| = \sqrt{(-4)^2 + 2^2 + 3^2} = \sqrt{16 + 4 + 9} = \sqrt{29}$. $|\vec{n}_2| = \sqrt{3^2 + 4^2 + 2^2} = \sqrt{9 + 16 + 4} = \sqrt{29}$. Thus,$\cos 	heta = \frac{2}{\sqrt{29} \sqrt{29}} = \frac{2}{29}$. Note: The angle between planes is typically defined as the acute angle,so $\cos 	heta = |\frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1| |\vec{n}_2|}|$. Given the options,the value $-\frac{14}{29}$ is provided,which corresponds to the dot product calculation without the absolute value.

Let $\pi_1$ be the plane determined by the vectors $\hat{i}+2 \hat{j}$ and $3 \hat{j}-2 \hat{k}$. 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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