Question #241630

(1) a) Give parametric equation (point-direction form) of the line which lies on both of the planes: x+y+z= 1 and -x+2y + 10z 2. What is the direction d of this line? b) Let ny and n₂ be the normal vectors to the two given planes. Without actual computation, describe the relationship between d and n₁ x n₂.


Expert's answer

a) n1=⟨1,1,1⟩,n2=⟨−1,2,10⟩n_1=\langle 1, 1, 1\rangle, n_2=\langle -1, 2, 10\rangle


n1×n2=∣ijk111−1210∣n_1\times n_2=\begin{vmatrix} i & j & k \\ 1 & 1 & 1 \\ -1 & 2 & 10 \\ \end{vmatrix}

=i∣11210∣−j∣11−110∣+k∣11−12∣=i\begin{vmatrix} 1 & 1 \\ 2 & 10 \end{vmatrix}-j\begin{vmatrix} 1 & 1 \\ -1 & 10 \end{vmatrix}+k\begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix}

=8i−11j+3k=8i-11j+3k

d=⟨8,−11,3⟩d=\langle 8, -11, 3\rangle

x+y+z=1x+y+z= 1−x+2y+10z=2-x+2y + 10z =2

x+y+z=1x+y+z= 13y+11z=33y+11z= 3

x1=0,y1=1,z1=0x_1=0, y_1=1, z_1=0


x2=8,y2=−10,z2=3x_2=8, y_2=-10, z_2=3

d=⟨x2−x1,y2−y1,z2−z1⟩d=\langle x_2-x_1, y_2-y_1, z_2-z_1\rangle

d=⟨8−0,−10−1,3−0⟩d=\langle 8-0, -10-1, 3-0\rangle

d=⟨8,−11,3⟩d=\langle 8, -11, 3\rangle

b)


d=n1×n2d=n_1\times n_2

d⊥n1,d⊥n2d\perp n_1, d\perp n_2


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