Calculate dist(Q,?) where, Q(2,1,1), and ? is determined by the normal vector to

Question

Calculate dist(Q,?) where, Q(2,1,1), and ? is determined by the normal vector to the plane n=(2,6,9)P and the point P(8,?3,1) lies on the plane.

Explanation / Answer

PLEASE RATE ME AND AWARD ME KARMA POINTS IF IT IS HELPFUL FOR YO 1. Find the equation of a sphere if one of its diameters has end points (1; 0; 5) and (5;¡4; 7). Solution: The length of the diameter is p (5 ¡ 1)2 + (¡4 ¡ 0)2 + (7 ¡ 5)2 = p 36 = 6, so the radius is 3. The centre is at the midpoint (1+5 2 ; 0¡4 2 ; 5+7 2 ) = (3;¡2; 6). Hence, the sphere is given as (x ¡ 3)2 + (y + 2)2 + (z ¡ 6)2 = 9 . 2. Find vector, parametric, and symmetric equations of the following lines. (a) the line passing through the points (3; 1; 1 2 ) and (4;¡3; 3) Solution: The vector between two points is ~v = h4 ¡ 3;¡3 ¡ 1; 3 ¡ 1 2 i = h1;¡4; 5 2 i. Hence the equation of the line is Vector form: ~r = ~r0 + t~v = h4;¡3; 3i + th1;¡4; 5 2 i = h4 + t;¡3 ¡ 4t; 3 + 5 2 ti Parametric form: x = 4 + t; y = ¡3 ¡ 4t; z = 3 + 5 2 t Symmetric from: Solving the parametric form for t gives x ¡ 4 = y+3 ¡4 = z¡3 5=2 (b) the line passing through the origin and perpendicular to the plane 2x ¡ 4y = 9 Solution: Perpendicular to the plane ) parallel to the normal vector ~n = h2;¡4; 0i. Hence Vector form: ~r = h0; 0; 0i + th2;¡4; 0i = h2t;¡4t; 0i Parametric from : x = 2t; y = ¡4t; z = 0 Symmetric form x 2 = y ¡4 ; z = 0 (c) the line lying on the planes x + y ¡ z = 2 and 3x ¡ 4y + 5z = 6 Solution: We can ¯nd the intersection (the line) of the two planes by solving z in terms of x, and in terms of y. (1) x + y ¡ z = 2 (2) 3x ¡ 4y + 5z = 6 Solve z in terms of y: 3 £ (1) ¡ (2) ) 7y ¡ 8z = 0 ) z = 7 8y Solve z in terms of x: 4 £ (1) + (2) ) 7x + z = 14 ) z = 14 ¡ 7x Hence, symmetric form: 14 ¡ 7x = 7 8y = z Set the symmetric form = t, we have parametric form: x = 14¡t 7 ; y = 8 7 t; z = t Vector form: ~r = h 14¡t 7 ; 8 7 t; ti 1 3. Find the equation of the following planes. (a) the plane passing through the points (¡1; 1;¡1), (1;¡1; 2), and (4; 0; 3) Solution: Name the points P(¡1; 1;¡1), Q(1;¡1; 2), and R(4; 0; 3). Set up two vectors: ~v1 = ¡! PQ = h1 + 1;¡1 ¡ 1; 2 + 1i = h2;¡2; 3i (1) ~v2 = ¡! PR = h5;¡1; 4i (2) Choose the normal vector ~n = ~v1 £ ~v2 = h¡5; 7; 8i. Hence the equation of the plane is ¡5(x + 1) + 7(y ¡ 1) + 4(z + 1) = 0 using point P. (b) the plane passing through the point (0; 1; 2) and containing the line x = y = z Solution: Name Q(0; 1; 2). The line can be represented as ~r = ht; t; ti, which crosses the point P(0; 0; 0) and is parallel to ~v = h1; 1; 1i. Set ~b = ¡! PQ = h0; 1; 2i. Choose ~n = ~v £~b = h1;¡2; 1i and hence the equation of the plane is x ¡ 2y + z = 0 using point P. (c) the plane containing the lines L1 : x = 1 + t; y = 2 ¡ t; z = 4t L2 : x = 2 ¡ s; y = 1 + 2s; z = 4 + s Solution: From L1 and L2, ~v1 = h1;¡1; 4i and ~v2 = h¡1; 2; 1i. Choose ~n = ~v1 £ ~v2 = h¡9;¡5; 1i. Since L1 crosses the point (1,2,0), the equation of the plane is ¡9(x ¡ 1) ¡ 5(y ¡ 2) + z = 0 . 4. Find the intersection of the line x = t, y = 2t, z = 3t, and the plane x + y + z = 1. Solution: Substitute the line into the plane: t + 2t + 3t = 1 ) t = 1 6 . Put t back to the line: x = 1 6 , y = 1 3 , z = 1 2 . Hence the intersection point is ( 1 6 ; 1 3 ; 1 2 ) . 5. Find the distance between the point (2; 8; 5) and the plane x ¡ 2y ¡ 2z = 1. Solution: Name Q(2; 8; 5). Choose any point on the plane, say a convenient one (x; 0; 0). So x¡2(0)¡2(0) = 1 ) x = 1 ) P(1; 0; 0). Then ~b = ¡! PQ = h1; 8; 5i. The normal vector of the plane is ~n = h1;¡2;¡2i. The distance between the plane and the point is given as 2 distance = ¯¯¯ proj~n ~b ¯¯¯ = j~n ¢~bj j~nj = j ¡ 25j j3j = 25 3 6. Show that the lines L1 : x ¡ 4 2 = y + 5 4 = z ¡ 1 ¡3 L2 : x ¡ 2 1 = y + 1 3 = z 2 are skew. Solution: Write the equation in parametric form. L1 : x = 2t + 4; y = 4t ¡ 5; z = ¡3t + 1 L2 : x = s + 2; y = 3s ¡ 1; z = 2s The lines are not parallel since the vectors ~v1 = h2; 4;¡3i and ~v2 = h1; 3; 2i are not parallel. Next we try to ¯nd intersection point by equating x, y, and z. (1) 2t + 4 = s + 2 (2) 4t ¡ 5 = 3s ¡ 1 (3) ¡ 3t + 1 = 2s (1) gives s = 2t+2. Substituting into (2) gives 4t¡5 = 3(2t+2)¡1 ) t = ¡5. Then s = ¡8. However, this contradicts with (3). So there is no solution for s and t. Since the two lines are neither parallel nor intersecting, they are skew lines. 7. Identify and sketch the following surfaces. (a) 4x2 + 9y2 + 36z2 = 36 Solution: xy-plane: 4x2 + 9y2 = 36 ellipse xz-plane: 4x2 + 36z2 = 36 ellipse yz-plane: 9y2 + 36z2 = 36 ellipse ) ellipsoid (b) 4z2 ¡ x2 ¡ y2 = 1 Solution: xy-plane: ¡x2 ¡ y2 = 1 nothing, try z = constants z = c: ¡x2 ¡ y2 = 1 ¡ 4c2 ) x2 + y2 = 4c2 ¡ 1 circles when 4c2 ¡ 1 > 0 xz-plane: 4z2 ¡ x2 = 1 hyperbola opening in z-direction yz-plane: 4z2 ¡ y2 = 1 hyperbola opening in z-direction ) hyperboloid of two sheets 3 (c) y2 = x2 + z2 Solution: xy-plane: y2 = x2 cross xz-plane: 0 = x2 + z2 point at origin, try y = constants y = c: c2 = x2 + z2 circles yz-plane: y2 = z2 cross ) cone (d) x2 + 4z2 ¡ y = 0 Solution: xy-plane: x2 ¡ y = 0 ) y = x2 parabola opening in +y-direction xz-plane: x2 + 4z2 = 0 point at origin, try y = constants y = c: x2 + 4z2 ¡ c = 0 ) x2 + 4z2 = c ellipses when c > 0 yz-plane: 4z2 ¡ y = 0 ) y = 4z2 parabola opening in +y-direction ) elliptic paraboloid (e) y2 + 9z2 = 9 Solution: x missing: cylinder along x-direction yz-plane: y2 + 9z2 = 9 ellipse ) elliptic cylinder (f) y = z2 ¡ x2 Solution: xy-plane: y = z2 parabola opening in +y-direction xz-plane: 0 = z2 ¡ x2 ) z2 = x2 cross, try y = constants y = c: c = z2 ¡ x2 hyperbola opening in z-direction when c > 0, in x-direction when c < 0 yz-plane: y = ¡x2 parabola opening in ¡y-direction ) hyperbolic paraboloid 8. Find the polar equation for the curve represented by the following Cartesian equation. (a) x = 4 Solution: x = 4 ) r cos µ = 4 ) r = 4 sec µ (b) x2 + y2 = ¡2x Solution: x2 + y2 = ¡2x ) r2 = ¡2r cos µ ) r = ¡2 cos µ 4 –2 0 2 –3 –2 –1 0 1 x y 2 3 –3 –2 –1 0 1 2 3 z Figure 1: Q7(a) –2 0 –1 2 1 –2 –1 0 1 x y –2 –1 0 1 2 z Figure 2: Q7(b) –2 0 –1 2 1 –2 –1 0 1 x y –2 –1 0 1 2 z Figure 3: Q7(c) –2 0 –1 2 1 –2 –1 0 1 x y –2 –1 0 1 2 z Figure 4: Q7(d) –2 0 2 x –3 –2 –1 0 1 2 3 y –3 –2 –1 0 1 2 3 z Figure 5: Q7(e) –1 0 1 –0.8 0 0.4 x y –0.8 –0.4 0 0.4 0.8 z Figure 6: Q7(f) (c) x2 ¡ y2 = 1 Solution: x2 ¡ y2 = 1 ) r2 cos2 µ ¡ r2 sin2 µ = 1 ) r2(cos2 µ ¡ sin2 µ) = 1 ) r2 cos 2µ = 1 ) r2 = sec 2µ ) r = § p sec 2µ 9. Sketch the curve of the following polar equations. (a) r = 5 (b) µ = 3¼ 4 (c) r = 2 sin µ (d) r = 3(1 ¡ cos µ) 10. (a) Change (3; ¼ 3 ; 1) from cylindrical to rectangular coordinates Solution: x = r cos µ = 3 cos ¼ 3 = 3 2 , y = r sin µ = 3 sin ¼ 3 = 3 p 3 2 , z = 1. Hence (x; y; z) = ( 3 2 ; 3 p 3 2 ; 1) (b) Change ( p 3; 1; 4) from rectangular to cylindrical coordinates Solution: r = p x2 + y2 = p 3 + 1 = 2, tan µ = y x = p1 3 ) µ = ¼ 6 in ¯rst quadrant, z = 4. Hence (r; µ; z) = (2; ¼ 6 ; 4) 5 –4 –2 2 4 –4 –2 2 4 Figure 7: Q9(a) –2 –1 0 1 2 –2 –1 1 2 Figure 8: Q9(b) –2 –1 0 1 2 –2 –1 1 2 Figure 9: Q9(c) –6 –4 –2 0 2 4 6 –6 –4 2 4 6 Figure 10: Q9(d) (c) Change ( p 3; 1; 2 p 3) from rectangular to spherical coordinates Solution: ½ = p x2 + y2 + z2 = p 3 + 1 + 12 = 4, tan µ = y x = p1 3 ) µ = ¼ 6 in ¯rst quadrant, Á = cos¡1 z ½ = cos¡1 2 p 3 4 = cos¡1 p 3 2 = ¼ 6 . Hence (½; µ; Á) = (4; ¼ 6 ; ¼ 6 ) (d) Change (4; ¼ 4 ; ¼ 3 ) from spherical to cylindrical coordinates Solution: r = ½ sin Á = 4 sin ¼ 3 = 2 p 3, µ = ¼ 4 , z = ½ cos Á = 4 cos ¼ 3 = 2. Hence (r; µ; z) = (2 p 3; ¼ 4 ; 2) 6

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