# Find the intercepts and sketch the plane 2 y0 4 2x 4

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2, b 3, c 4, x 0 see that an equation of the plane is (0, 0, 3) (6, 0, 0) 4, and z0 2, y0 1 in Equation 7, we 2 3y or 4 4z 1 0 2x 2x (0, 4, 0) 1 with normal vector 2, 3, 4 . Find the intercepts and sketch the plane. n 3y 4z 12 y To ﬁnd the x-intercept we set y z 0 in this equation and obtain x 6. Similarly, the y-intercept is 4 and the z-intercept is 3. This enables us to sketch the portion of the plane that lies in the ﬁrst octant (see Figure 7). x FIGURE 7 By collecting terms in Equation 7 as we did in Example 4, we can rewrite the equation of a plane as ax 8 by cz d 0 where d a x 0 b y0 cz0 . Equation 8 is called a linear equation in x, y, and z. Conversely, it can be shown that if a, b, and c are not all 0, then the linear equation (8) represents a plane with normal vector a, b, c . (See Exercise 73.) |||| Figure 8 shows the portion of the plane in Example 5 that is enclosed by triangle PQR. EXAMPLE 5 Find an equation of the plane that passes through the points P 1, 3, 2 , Q 3, 1, 6 , and R 5, 2, 0 . l Q(3, -1, 6) a P(1, 3, 2) 2, 4, 4 b 4, 1, Since both a and b lie in the plane, their cross product a and can be taken as the normal vector. Thus y n x a i 2 4 b R(5, 2, 0) FIGURE 8 l SOLUTION The vectors a and b corresponding to PQ and PR are z j 4 1 k 4 2 2 b is orthogonal to the plane 12 i 20 j 14 k With the point P 1, 3, 2 and the normal vector n, an equation of the plane is 12 x or 1 20 y 3 14 z 2 0 6x 10 y 7z 50 5E-13(pp 858-867) 1/18/06 11:29 AM Page 863 SECTION 13.5 EQUATIONS OF LINES AND PLANES E XAMPLE 6 Find the point at which the line with parametric equations x y 4 t, z 5 t intersects the plane 4 x 2z 5y 2 ❙❙❙❙ 863 3 t, 18. SOLUTION We substitute the expressions for x, y, and z from the parametric equations into the equation of the plane: 42 3t 5 4t 25 t 18 This simpliﬁes to 10 t 20, so t 2. Therefore, the point of intersection occurs when the parameter value is t 2. Then x 2 3 2 4, y 42 8, z 5 2 3 and so the point of intersection is 4, 8, 3 . n™ ¨ n¡ Two planes are parallel if their normal vectors are parall...
View Full Document

## This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas at Austin.

Ask a homework question - tutors are online