Compute the cosine of the angle between the plane through P = (6,0,0), Q = (0, 5, 0), and R = (0,0,9) and the yz-plane,defined as the angle between their normal vectors.(Use symbolic notation and fractions where needed.)cos()- The plane + += 1 intersects the x-, y-, and z-axes in points P, Q, R. Find the area of the triangle APQR.SAPOR

Answered: Compute the cosine of the angle between…

Algebra and Trigonometry (MindTap Course List)

4th Edition

ISBN:9781305071742

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter9: Vectors In Two And Three Dimensions

Section9.6: Equations Of Lines And Planes

Problem 2E

See similar textbooks

Related questions

Question

Compute the cosine of the angle between the plane through P = (6,0,0), Q = (0, 5, 0), and R = (0,0,9) and the yz-plane,
defined as the angle between their normal vectors.
(Use symbolic notation and fractions where needed.)
cos()-

The plane + +
= 1 intersects the x-, y-, and z-axes in points P, Q, R. Find the area of the triangle APQR.
SAPOR

Expert Solution

Step 1

(a). Given points are $P (6, 0, 0), Q (0, 5, 0)$ and $R (0, 0, 9)$ .

We know that equation of plane through $(a, 0, 0), (0, b, 0)$ and $(0, 0, c)$ is

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

Therefore equation of plane through $P (6, 0, 0), Q (0, 5, 0)$ and $(0, 0, 9)$ will be

$\begin{array}{rcl} \frac{x}{6} + \frac{y}{5} + \frac{z}{9} & = & 1 \\ 15 x + 18 y + 10 z - 90 & = & 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) \end{array}$

Hence normal vector to the above plane is $\vec{n_{1}} = 15 \hat{i} + 18 \hat{j} + 10 \hat{k}$ .

The equation of $y z -$ plane is $x = 0$ . Therefore the normal vector to the $y z -$ plane is $\vec{n_{2}} = \hat{i}$ .

Therefore cosine of the angle between above two plane is given by:
$\begin{array}{rcl} \cos θ & = & |\frac{\vec{n_{1}} . \vec{n_{2}}}{|\vec{n_{1}}| |{\vec{n}}_{2}|}| \\ = & |\frac{(15 \hat{i} + 18 \hat{j} + 10 \hat{k}) . (\hat{i})}{|15 \hat{i} + 18 \hat{j} + 10 \hat{k}| |\hat{i}|}| \\ = & |\frac{15}{\sqrt{15^{2} + 18^{2} + 10^{2}} \sqrt{1^{2}}}| \\ = & \frac{15}{\sqrt{649}} \end{array}$

Therefore $\cos θ = \frac{15}{\sqrt{649}}$

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Projection$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Recommended textbooks for you

Algebra and Trigonometry (MindTap Course List)

Algebra

ISBN:

9781305071742

Author:

James Stewart, Lothar Redlin, Saleem Watson

Publisher:

Cengage Learning