# X 3 y 2 z 7 0 2x4 4 x 8 y 20 z 4 0 11 y 22 z 11 0 y 2

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x + 3 y 2 z + 7 = 0 [2]x4 4 x 8 y + 20 z 4 = 0 11 y 22 z + 11 = 0 y 2 z + 1 = 0 y =− 1 + 2 z Now substituting the value of y in Equation [2]
x 2 ( 1 + 2 z ) + 5 z 1 = 0 x + 2 4 z + 5 z 1 = 0 x =− 1 z Now let z = t and writing x and y in terms of t to form a parametric equation. x =− 1 t y =− 1 + 2 t z = t Writing it in vector form, ´ w = ( 1, 1,0 ) + t ( 1,2,1 ) 7/7 marks 16. 4 x + y 9 z = 0 [1] x + 2 y + 3 z = 0 [2] 2 x 3 y 5 = 0 [3] The normal vectors for the planes are n 1 = ( 4,1, 9 ) ,n 2 = ( 1,2,3 ) n 3 =( 2, 3,0 ) Calculate
´ n 1 ( ´ n 2 × ´ n 3 ) = ( 4,1, 9 ) . ( ( 1,2,3 ) × ( 2, 3,0 ) ) ¿ ( 4,1, 9 ) . ( 2 ( 0 ) 3 ( 3 ) , 3 ( 2 ) 1 ( 0 ) , 1 ( 3 ) 2 ( 2 ) ) ¿ ( 4,1, 9 ) . ( 9,6, 7 ) ¿ 4 ( 9 ) + 1 ( 6 ) + ( 9 ) ( 7 ) ¿ 105 0 Hence the planes intersect at a point. Using Equation [2] and [3] to eliminate x [2]x2 2 x + 4 y + 6 z = 0 [3] 2 x 3 y 5 = 0 7 y + 6 z + 5 = 0 [4] Using equation [1] and [2] to eliminate x [1] 4 x + y 9 z = 0 [2]x4 4 x + 8 y + 12 z = 0 7 y 21 z = 0
y =− 3 z Substituting this value of y in Equation [4] 7 ( 3 z ) + 6 z + 5 = 0 15 z + 5 = 0 z = 1 3 y =− 3 ( 1 3 ) =− 1 Substituting the values of y and z in Equation [3] 2 x 3 ( 1 ) 5 = 0 2 x = 2 x = 1 Hence the point of intersection is ( 1, 1, 1 3 )
4/4 marks 17. P ( 4,2,6 ) 2 x 3 y + z 8 = 0 The normal vector to the plane is ´ n = ( 2, 3,1 ) Let’s find another point R on the plane, For x = 0 & y = 0 2 ( 0 ) 3 ( 0 ) + z 8 = 0 z = 8 Now we have point R ( 0,0,8 ) ´ PR = R P = ( 0,0,8 ) ( 4,2,6 ) ¿ ( 4, 2,2 ) Now we can use a projection of ´ PR on ´ n to find the distance | Pro j ´ n ´ PR | = | ´ PR . ´ n ´ n. ´ n | | ´ n | ¿ | ( 4, 2,2 ) . ( 2, 3,1 ) ( 2, 3,1 ) ( 2, 3,1 ) | | 2, 3,1 | ¿ | 4 ( 2 ) + ( 2 ) ( 3 ) + 2 ( 1 ) 2 ( 2 ) + ( 3 ) ( 3 ) + 1 ( 1 ) | 2 2 + ( 3 ) 2 + 1 2 ¿ 16 14 14 ¿ 4.28 units
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