6. At point P(-10,0) in cartesian coordinates, given A = 3P-20 and B = @ + 32. The result of A B (A) 6. (9 9. (B) -6 (E) 3. (D) -9 (F) 7. Given the vector field A = 3rt - 2sineô. At point P(1,30°, 0º), the vector perpendicular to A and tangential to the cone e = 30° is (A) 3f +30 (9 30 (B) 3P-30 (E) 28 + 30 (D) 3P (F) No answer is correct 8. The distance between the two lines F, (t)= &+ tỷ, (S)= -28 + s2 is (A) V3 (B) 4 (C) 23 (D) 2. (E) 3 (F) 9. The intersection point of the plane 4x - 2y + z = 1 and the line F(t)= & + (1+t)9 is 3. (B) (1,5.0) 5. (A) (-1,-2,1) (D) (1,5.0) -3 (1. -6) (E) (0..2) (C) (1,1,0) (F) of the surface 0 = 45° with the sw
6. At point P(-10,0) in cartesian coordinates, given A = 3P-20 and B = @ + 32. The result of A B (A) 6. (9 9. (B) -6 (E) 3. (D) -9 (F) 7. Given the vector field A = 3rt - 2sineô. At point P(1,30°, 0º), the vector perpendicular to A and tangential to the cone e = 30° is (A) 3f +30 (9 30 (B) 3P-30 (E) 28 + 30 (D) 3P (F) No answer is correct 8. The distance between the two lines F, (t)= &+ tỷ, (S)= -28 + s2 is (A) V3 (B) 4 (C) 23 (D) 2. (E) 3 (F) 9. The intersection point of the plane 4x - 2y + z = 1 and the line F(t)= & + (1+t)9 is 3. (B) (1,5.0) 5. (A) (-1,-2,1) (D) (1,5.0) -3 (1. -6) (E) (0..2) (C) (1,1,0) (F) of the surface 0 = 45° with the sw
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
siiirr please solve question7
Transcribed Image Text:6.
At point P(-10,0) in cartesian coordinates, given A = 3P-20 and B = @ + 32. The result of A B
(A)
6.
(9
9.
(B)
-6
(E)
(D)
-9
(F)
-3
7.
Given the vector field A = 3rt- 2sineê. At point P(1,30°, 0º), the vector perpendicular to A and
tangential to the cone 0 = 30° is
(A)
3f +30
30
(B)
3P-30
(E)
28 + 30
(D)
3P
(F)
No answer is correct
8. The distance between the two lines , (t)= + tỷ, F(s)= -2£ + s2 Is
(A)
V3
(B)
4
()
23
(D)
2.
(E)
3
(F)
9.
The intersection point of the plane 4x -2y +z = 1 and the line F(t)= & + (1+t)9 is
3
(B)
(1,5.0)
(A)
(-1,-2,1)
(D)
(1.5.0)
(1.-6)
(E) (0..2)
(C)
(1,1,0)
(F)
of the surface 0 = 45° with the sw
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