Find the angle of intersection of the planes in degrees. Round to a tenth of a degree.
$$\begin{array} { l }
3 x - 2 y + 4 z = 3 \
5 x - 3 y + 3 z = - 6
\end{array}$$
A) $- 0.8 ^ { \circ }$
B) $69.1 ^ { \circ }$
C) $20.9 ^ { \circ }$
D) $43.1 ^ { \circ }$
E) $0.4 ^ { \circ }$

Question

Find the angle of intersection of the planes in degrees. Round to a tenth of a degree.
$$\begin{array} { l } 
3 x - 2 y + 4 z = 3 \
5 x - 3 y + 3 z = - 6
\end{array}$$
A) $- 0.8 ^ { \circ }$
B) $69.1 ^ { \circ }$
C) $20.9 ^ { \circ }$
D) $43.1 ^ { \circ }$
E) $0.4 ^ { \circ }$

Quizplus · Accepted Answer

We need to find the angle formed by the normal vectors of the two planes. We can find the normal vectors by looking at the coefficients of each variable. The normal vector of the first plane is $\begin{pmatrix} 3 \ -2 \ 4 \end{pmatrix}$ and the normal vector of the second plane is $\begin{pmatrix} 5 \ -3 \ 3 \end{pmatrix}$. We can find the angle between these vectors using the dot product formula:

$$\cos	heta = \frac{\begin{pmatrix} 3 \ -2 \ 4 \end{pmatrix} \cdot \begin{pmatrix} 5 \ -3 \ 3 \end{pmatrix}}{\|\begin{pmatrix} 3 \ -2 \ 4 \end{pmatrix}\| \|\begin{pmatrix} 5 \ -3 \ 3 \end{pmatrix}\|}$$

Simplifying, we get:

$$\cos	heta = \frac{23}{\sqrt{29}\sqrt{43}}$$

Taking the inverse cosine of both sides and rounding to the nearest tenth of a degree, we get:

$$	heta \approx 20.9 ^ { \circ }$$

Therefore, the answer is choice C.

Find the Angle of Intersection of the Planes in Degrees 3x−2y+4z=35x−3y+3z=−6\begin{array} { l } 3 x - 2 y + 4 z = 3 \\5 x - 3 y + 3 z = - 6\end{array}3x−2y+4z=35x−3y+3z=−6​

Find the Angle of Intersection of the Planes in Degrees $\begin{array} { l } 3 x - 2 y + 4 z = 3 \\5 x - 3 y + 3 z = - 6\end{array}$