The plane r→ = si^ + j^ - 

Subject

Mathematics

Class

JEE Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

31.

If the vectors a = 2i^ + j^ + 4k^, b = 4i^ - 2j^ + 3k^ and c = 2i^ - 3j^ - λk^ are coplanar, then the value of λ is equal to

  • 2

  • 1

  • 3

  • - 1


32.

Let A(1, - 1, 2) and B (2, 3, - 1) be two points. If a point P divides AB internally in the ratio 2 : 3, then the position vector of P is

  • 15i^ + j^ + k^

  • 13i^ +  6j^ + k^

  • 13i^ + j^ + k^

  • 157i^ + 3j^ + 4k^


33.

If the scalar product of the vector i^ + j^ + 2k^ with the unit vector along mi^ + 2j^ + 3k^ is equal to 2, then one of the values of m is

  • 3

  • 4

  • 5

  • 6


34.

A plane makes intercepts a, b, cat A, B, C on the coordinate axes respectively. If the centroid of the ABC is at (3, 2, 1), then the equation of the plane is

  • x + 2y + 3z = 9

  • 2x - 3y - 6z = 18

  • 2x + 3y + 6z = 18

  • 2x + y + 6z = 18


Advertisement
35.

If the plane 3x + y + 2z + 6 = 0 is parallel to the line 3x - 12b = 3 - y = z - 1a, then the avlue of 3a + 3b is

  • 12

  • 32

  • 3

  • 4


36.

The equation of the line passing through the point (3, 0,- 4) and perpendicular to the plane 2x - 3y + 5z - 7 = 0 is

  • x - 23 = y- 3 = z + 45

  • x - 32 = y- 3 = z - 45

  • x - 23 = - y3 = z + 45

  • x + 32 = y3 = z - 45


Advertisement

37.

The plane r = si^ + j^ - 4k^ + t3i^ +4 j^ - 4k^ + 1 - t2i^ - 7j^ - 3k^ is parallel to the line

  • r = - i^ + j^ - k^ + t- i^ - 2j^ + 4k^

  • r = - i^ + j^ - k^ + t i^ - 2j^ + 4k^

  • r = i^ + j^ - k^ + t- i^ - 4j^ + 7k^

  • r = - i^ + j^ - 3k^ + t2i^ + 6j^ - 8k^


A.

r = - i^ + j^ - k^ + t- i^ - 2j^ + 4k^

Given plane is

    r = si^ + j^ - 4k^ + t3i^ + 4j^ - 4k^ + 1 - t2i^ - 7j^ - 3k^ r = 2i^ - 7j^ + 3k^ + si^ + 2j^ - 4k^ + ti^ + 11j^ - k^Comparing it with the equation of plane        r = a + λb + μc, we get       b = i^ + 2j^ - 4k^and c = i^ + 11j^ - k^Now, b × c = i^j^k^12- 4111- 1                     = 42i^ - 3j^ + 9k^ Parametric form of plane isr . b × c = a . b × cr . 42i^ - 3j^ + 9k^     = 2i^ - 7j^ + 3k^ . 42i^ - 3j^ + 9k^which is of the form r . r = d  r = 42i^ - 3j^ + 9k^Now, the line given in option (a) isr = - i^ + j^ + k^ + t- i^ - 2j^ + 4k^Comparing it with r = p + tq, we getq = - i^ - 2j^ + 4k^Since, q . n = - i^ - 2j^ + 4k^ . 42i^ - 3j^ + 9k^                    = - 42 + 6 + 36 = 0

Hence, the line given in option (a) is parallel to the given plane

 


Advertisement
38.

The distance between the line r = 2i^ + 2j^ - k^ + λ2i^ + j^ - 2k^ and the plane r . i^ + 2j^ + 2k^ = 10 is equal to

  • 5

  • 4

  • 3

  • 2


Advertisement
39.

Equation of the plane passing through t intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0 and the point (1, 1, 1)

  • 20x + 23y + 26z - 69 = 0

  • 31x + 45y + 49z + 52 = 0

  • 8x + 5y + 2z - 69 = 0

  • 4x + 5y + 6z - 7 = 0


40.

The equation of the plane containing the line x - 12 = y + 1- 1 = z3 and x2 = y - 2- 1 = z + 13 is

  • 8x - y + 5z - 8 = 0

  • 8x + y - 5z - 7 = 0

  • x - 8y + 3z + 6 = 0

  • 8x + y - 5z + 7 = 0


Advertisement