Find the angle between the planes whose vecto | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equations of the given planes are $\vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5$ and $\vec{r} \cdot (3 \hat{i} - 3 \hat{j} + 5 \hat{k

Vedclass · Accepted Answer

$\cos Q = \frac{15}{\sqrt{731}}$

Vedclass · Answer

(A) The equations of the given planes are $\vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5$ and $\vec{r} \cdot (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3$.The normal vectors to the planes are $\vec{n}_{1} = 2 \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $\vec{n}_{2} = 3 \hat{i} - 3 \hat{j} + 5 \hat{k}$.The angle $Q$ between the two planes is given by the formula:$\cos Q = \left| \frac{\vec{n}_{1} \cdot \vec{n}_{2}}{|\vec{n}_{1}| |\vec{n}_{2}|} \right|$First,calculate the dot product $\vec{n}_{1} \cdot \vec{n}_{2}$:$\vec{n}_{1} \cdot \vec{n}_{2} = (2)(3) + (2)(-3) + (-3)(5) = 6 - 6 - 15 = -15$.Next,calculate the magnitudes $|\vec{n}_{1}|$ and $|\vec{n}_{2}|$:$|\vec{n}_{1}| = \sqrt{2^2 + 2^2 + (-3)^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$.$|\vec{n}_{2}| = \sqrt{3^2 + (-3)^2 + 5^2} = \sqrt{9 + 9 + 25} = \sqrt{43}$.Now,substitute these values into the formula:$\cos Q = \left| \frac{-15}{\sqrt{17} \cdot \sqrt{43}} \right| = \frac{15}{\sqrt{731}}$.Thus,the angle $Q$ is $\cos^{-1} \left( \frac{15}{\sqrt{731}} \right)$.

Find the angle between the planes whose vector equations are $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5$ and $\vec{r} \cdot(3 \hat{i}-3 \hat{j}+5 \hat{k})=3$.

Similar Questions

If a variable plane,at a distance of $3 \ units$ from the origin,intersects the coordinate axes at $A, B$,and $C$,then the locus of the centroid of $\Delta ABC$ is

Find the equation of a plane,given that the foot of the perpendicular drawn to the plane from the origin is $(2, 1, 2)$.

The points $A(-1, 3, 0)$,$B(2, 2, 1)$,and $C(1, 1, 3)$ determine a plane. The distance from the plane to the point $D(5, 7, 8)$ is

Let $Q$ be the foot of the perpendicular from the origin to the plane $4x - 3y + z + 13 = 0$ and $R$ be a point $(-1, 1, -6)$ on the plane. Then the length $QR$ is

In the following cases,determine whether the given planes are parallel or perpendicular,and in case they are neither,find the angles between them.
$2x + y + 3z - 2 = 0$ and $x - 2y + 5 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module