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**Q.1. A vector lies in the plane of the vectors If bisects the angle between then (2020)(1) (2) (3) (4) Ans.** (3)

We have,

On comparing with

We get Î± = 4 and Î² = 4

Therefore,

Now, again consider

On comparing with we get

Therefore,

(1)

(2)

(3)

(4)

Ans.

Given

...(1)

Taking square of both sides, we get

(since are unit vectors)

Now,

(1) -3/2

(2) 1/2

(3) -1/2

(4) -1

Ans.

Given

...(1)

Now,

and

Substituting these values in Eq. (1), we get

Hence,

Ans.

We have

Hence, the required projection is

Ans.

The vectors are coplanar vectors, then

Therefore, ...(1)

...(2)

...(3)

Now, ...(4)

...(5)

Hence,

Ans.

Given,

Now, is perpendicular to the vector then

Hence,

Ans.

The given lines must be intersecting

Therefore,

(2s - 1, 3s + 3, 8s - 1) = (2t - 3t, t - 2, Î»t + 1)

Now, 2s - 1 = 2t - 3 ...(1)

3s + 3 = t - 2 ...(2)

8s - 1 = Î»t + 1 ...(3)

On solving above equations, we get

t = -1, s = -2 and Î» = 18

Distance of plane contains given lines from given plane is same as distance between point (â€“3, â€“2, 1) from given plane. So,**Q.8. Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then, the image of R in the plane P is (2020)(1) (6, 5, 2)(2) (6, 5, âˆ’2)(3) (4, 3, 2)(4) (3, 4, âˆ’2)Ans.** (2)

The equation of the plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) is

â‡’ (x - 2)(1) - (y - 1)(2 - 3) + z(-2) = 0

â‡’ x + y - 2z = 3

Let I and F are respectively image and foot of perpendicular of point R in the plane. So, the equation of RI is

(say)

â‡’ x = k + 2, y = k + 1 and z = -2k + 6

Hence, the coordinates of point F is

Point F lies in the plane, then

â‡’ 3k = 12 â‡’ k = 4

Hence, the image I is (6, 5, -2).

Ans.

We have,

Direction ratio of BP

Direction ratio of AP =

Since AP and BP are perpendicular, then

(1) 2âˆš30

(2) 7/2 (âˆš30)

(3) 3âˆš30

(4) 3

Ans.

The equations of the lines are

...(1)

...(2)

The distance between the lines is given by

Now,

Hence,

(1)

(2)

(3) 5/7

(4)

Ans.

The volume of parallelepiped is 1 cu. unit. Therefore,

â‡’ Î» = 2, 4

Now,

(1) (1, 1, 1)

(2) (1, -1, 1)

(3) (-1, -1, 1)

(4) (-1, -1, -1)

Ans.

Direction ratios of normal to the plane is

<1 + (7/3), 2 + (4/3),3 + (1/3)> = <1, 1, 1>

The coordinates of mid- point of image and point is

The equation of plane is

â‡’ x + y + z = 1

Hence, the point (1, âˆ’1, 1) lies on the plane.

x + 4y - 2z =1

x + 7y - 5z = Î²

x + 5y - Î±z = 5

is a line in then Î± + Î² is equal to (2020)

(1) 0

(2) 10

(3) 2

(4) âˆ’10

Ans.

We have

=0

â‡’ (7Î± + 25) - (4Î± + 10) + (-20 + 14) = 0

â‡’ 3Î± + 9 = 0 â‡’ Î± = -3

and

Hence, Î± + Î² = -3 + 13 = 10

(1) 19/2

(2) 9

(3) 8

(4) 17/2

Ans.

If is perpendicular to is equal to: (2019)

(1) âˆš32

(2) 6

(3) âˆš22

(4) 4

Ans.

Projection of

According to question

â‡’ b

Since, is perpendicular to

Hence

â‡’ 8 + 5b

From (1) and (2),

(1) (1, 3, 1)

(2) (-1/2, 4, 0)

(3) (1/2, 4, -2)

(4) (1, 5, 1)

Ans.

âˆ´ 3 - Î»

...(2)

Since, (-1/2, 4, 0) satisfies equation (1) and (2). Hence, one of possible value of

(1) -4

(2) -3

(3) 4

(4) 3

Ans.

Let are collinear for same k

i.e.,

But are non-collinear, then

â‡’

Î» = -4

(1)

(2)

(3)

(4)

Ans.

are coplanar

For (Rejected)

(1) 4

(2) 3

(3) 2

(4) 1

Ans.

Since, the angle bisector of acute angle between OA and OB would be y = x

Since, the distance of C from bisector = 3/âˆš2

Hence, the sum of all possible value of Î² = 2 + (-1) = 1

(2) 0

(3) 1

(4) 2

Ans.

âˆµ Three vectors and are copalnar.

Therefore, sum of all real values = 1 - 2 = -1

(1) 30Â°

(2) 90Â°

(3) 60Â°

(4) 45Â°

Ans.

Since, are three unit vectors

âˆ´

Then,

âˆ´

â‡’ Î² = 60Â° and Î± = 90Â°

Hence,

(1) âˆš3/2

(2) âˆš6

(3) 3âˆš6

(4)

Ans.

Let

âˆ´ Vector perpendicular to

Now, projection of vector is

(1)

(2)

(3)

(4)

Ans.

Given

Now,

(1)

(2)

(3)

(4)

Ans.

...(1)

Since, is perpendicular to

Cross product with in equation (1)

(1) 5Ï€/6

(2) Ï€/4

(3) 5Ï€/12

(4) 2Ï€/3

Ans.

Let cos Î±, cos Î², cos Î³ be direction cosines of a.

(1) 7

(2) 4âˆš3

(3) 6

(4) 2âˆš13

Ans.

Equation of the line is

Let the point M on the line

Direction ratio's of PM is (6Î» + 3, 3Î» + 1, - 4Î»- 10)

â‡’ (6Î» + 3) (6) + (3Î» + 1) (3) + (-4Î».- 10) (- 4) = 0

â‡’ Î» = - 1 â‡’ M = (- 4, 0, 0)

(1)

(2)

(3)

(4)

Ans.

Let vector be

Given that magnitude of the vector is 12.

(1) is singleton

(2) is empty

(3) contains exactly two positive numbers

(4) contains exactly two numbers only one of which is positive

Ans.

Let, three vectors are coplanar, then

âˆµ no real value of â€˜Î±â€™ exist,

âˆ´ set S is an empty set.

(1)

(2)

(3)

(4)

Ans.

Let any point on the intersecting line

is (-3Î» - 1, 2Î» + 3, -Î» + 2)

Since, the above point lies on a line which passes through the point (-4, 3, 1)

Then, direction ratio of the required line

Since, line is parallel to the plane

x + 2y - z - 5 = 0

Then, perpendicular vector to the line is

Now (-3Î» + 3)(1) + (2Î»)(2) + (-Î» + 1)(-1) = 0

â‡’ Î» = - 1

Now direction ratio of the required line = <6, - 2, 2 > or <3,-1, 1>

Hence required equation of the line is

(1) (-3, 0, -1)

(2) (-3, 1, 1)

(3) (3, 3, -1)

(4) (3, 2, 1)

Ans.

Since, equation of plane through intersection of planes

But, the above plane is parallel to y-axis then

â‡’ Î» = -3

Hence, the equation of required plane is

-x - 4z + 7 = 0

â‡’ x + 4z - 7 = 0

Therefore, (3, 2, 1) the passes through the point.

(1) x - 2y + z = 0

(2) 3x + 2y - 3z = 0

(3) x + 2y - 2z = 0

(4) 5x + 2y - 4z = 0

Ans.

Let the direction ratios of the plane containing lines

âˆ´ Direction ratio of plane = <-8, 1, 10>.

Let the direction ratio of required plane is <

Then-81 + m + 10n = 0 ...(1)

and 2l + 3m + 4n = 0 ...(2)

From (1) and (2),

âˆ´ D.R.s are <1, -2, 1>

âˆ´ Equation of plane: x - 2y + z = 0**Q.32. The plane passing through the point (4, -1, 2) and parallel to the lines also passes through the point: (2019)(1) (1, 1, -1)(2) (1, 1, 1)(3) (-1, -1,-1)(4) (-1, -1, 1)Ans. **(2)

Equation of required plane is

(x - 4)(-3 -4) - (y + 1)(9 - 2) + (z - 2) (6 + 1) = 0

âˆµ point (1, 1, 1) satisfies this equation

âˆ´ point (1, 1, 1) lies on the plane

(1) 1/4

(2) 1/8

(3) 1/2

(4) -1/4

Ans.

âˆµ A be a point on given line.

âˆ´ Position vector of A

Equation of plane is: x - 4y + 3z = 1

âˆµ is parallel to this plane.

â‡’ Î¼ = 1/4

(1) (-2, 3, 5)

(2) (4, -1, 7)

(3) (2, 1, 3)

(4) (4, 1, -2)

Ans.

Q(0, 2, 5)

Since, direction ratios of normal to the plane is

Then, equation of the plane is

(x - 0)6 + (y - 2)10 + (z - 5)2 = 0

3x + 5y - 10 + z - 5 = 0

3x + 5y + z = 15 ...(1)

Since, plane (1) satisfies the point (4, 1, -2)

Hence, required point is (4, 1, -2)

(1)

(2)

(3)

(4)

Ans.

Let any point on the line be A(2Î» + 4, 2Î» + 5, Î», + 3) which lies on the plane x +y + z = 2

â‡’ 2Î» + 4 + 2Î» + 5 + Î» + 3 = 2

â‡’ 5Î» = -10 â‡’ Î» = - 2

Then, the point of intersection is (0, 1, 1)

which lies on the line

(1) (2, 2, 0)

(2) (-2, 2, 2)

(3) (0, -2, 2)

(4) (2, 0, -2)

Ans.

Let normal to the required plane is

â‡’ is perpendicular to both vector

â‡’ Equation of the required plane is

â‡’ (x - 3)(-8) + (y + 2) x 8 + (z - 1) x 8 = 0

â‡’ (x - 3) (-1) + (y + 2) x 1 + (z - 1) x 1 = 0

â‡’ x - 3 - y - 2 - z + 1 =0

âˆµ x - y - z = 4 passes through (2, 0, -2)

âˆ´ plane contains (2, 0, -2).

(1) 2, -1, 1

(2) 2, âˆš2, -âˆš2

(3) âˆš2, 1, -1

(4) 2âˆš3, 1, -1

Ans.

Let the d.râ€™s of the normal be (a, b, c)

Equation of the plane is

a(x - 0) + b(y + 1) + c(z - 0) = 0

âˆµ It passes through (0, 0, 1)

âˆ´ b + c

Also

âˆ´ The d.râ€™s are âˆš2, 1, -1 or 2, âˆš2, -âˆš2

(1) (2, -4, -7)

(2) (2, 4, 7)

(3) (2, -4, 7)

(4) (-2, 4, 7)

Ans.

Let the coordinate of P with respect to line

and coordinate of P w.r.t.

From above equation : Î» = -1, Î¼ = 1 .

âˆ´ Coordinate of point of intersection R = (2, -4, 1).

Image of R w.r.t. xy plane = (2, -4, -7).

(1) 12

(2) 7

(3) 5

(4) 17

Ans.

Let the normal to the required plane is n, then

âˆ´ Equation of the plane

(x - 3) x 20 + (y - 4) x 8 + (z - 2) x (-12) = 0

5x - 15 + 2y - 8 - 3z + 6 = 0

5x + 2y - 3z - 17 = 0 ...(1)

Since, equation of plane (1) passes through (2, Î±, Î²), then

(1) 11âˆš6

(2) 11/âˆš6

(3) 11

(4) 6/âˆš11

Ans.

âˆµ plane containing both lines.

âˆ´ D.R. of plane

Now, equation of plane is,

7(x - 1) - 14(y - 4) + 7 (z + 4) = 0

â‡’ x - 1 - 2y + 8 + z + 4 = 0

â‡’ x - 2y + z + 11 = 0

Hence, distance from (0, 0, 0) to the plane,

(1)

(2)

(3) -3/5

(4) -5/3

Ans.

Let angle between line and plane is Î¸, then

Since,

Then,

(1) {âˆš3}

(2) {âˆš3, -âˆš3}

(3) {1, -1}

(4) {3, -3}

Ans.

Let D(-1, -1, 1) lie on same plane, then

â‡’ (Î»

â‡’ (3 - Î»

Hence, S = {-âˆš3, âˆš3}

(1) x- 3y- 2z = -2

(2) 2x - z = 2

(3) x - y - z = 0

(4) x + 3y + z = 4

Ans.

Let the equation of required plane be;

(2x - y - 4) + Î»(y + 2z - 4) = 0

âˆ´ This plane passes through the point (1, 1, 0) then (2 - 1 - 4) + Î»(1 + 0 - 4) = 0

â‡’ Î» = -1

Then, equation of required plane is,

(2x - y - 4) - (y + 2z - 4) = 0

â‡’ 2x - 2y â€” 2z = 0 â‡’ x - y - z = 0

(1) greater than 3 but less than 4

(2) less than 2

(3) greater than 2 but less than 3

(4) greater than 4

Ans.

Let P be the foot of perpendicular from point T (2, -1, 4) on the given line. So P can be assumed as P(10Î» - 3, -7Î» + 2, Î»)

DR's of TP is proportional to 10Î» - 5, - 7Î» + 3, Î» - 4

âˆµ TP and given line are perpendicular, so

10(10Î» - 5) -7 (- 7Î» + 3)+ 1(Î» - 4) = 0

Hence, the length of perpendicular is greater than 3 but less than 4.

(1)

(2)

(3)

(4)

Ans.

Equation of the plane passing through the line of intersection of x + y + z = 1 and 2x + 3y + 4z = 5 is (2x + 3y + 4z - 5) + Î» (x + y + z - 1) = 0

âˆµ plane (i) is perpendicular to the plane x-y + z = 0

âˆ´ (2 + Î»)(1) + (3 + Î»)(- 1) + (4 + Î»)(1) = 0

2 + Î» - 3 - Î» + 4 + Î» = 0

â‡’ Î» = -3

Hence, equation of required plane is -x + z- 2 = 0 or x - z + 2 = 0

(1) 2âˆš14

(2) 2âˆš21

(3) 6

(4) âˆš53

Ans.

Here, P, Q, R are collinear

Now, or =

(1) âˆš5 / 2

(2) 2 / âˆš5

(3) 9/2

(4) 7/2

Ans.

Let point on line be P (2k + 1, 3k - 1, 4k + 2)

Since, point P lies on the plane x + 2y + 3z = 15

âˆ´ 2k + 1 + 6k - 2 + 12k + 6= 15

â‡’ k = 1/2

Then the distance of the point P from the origin is

(1) (-âˆš2, 1, -4)

(2) (-âˆš2, -1, 4)

(3) (-âˆš2, -1, -4)

(4) (âˆš2, 1, 4)

Ans.

Let the required plane passing through the points (0,-1, 0) and (0, 0, 1) be and the given plane is y - z + 5 = 0

Then, the equation of plane is

Then the point (âˆš2, 1, 4) satisfies the equation of plane -âˆš2x - y + z = 1

(1) 5âˆš17

(2) 2âˆš34

(3) 6

(4) âˆš34

Ans.

Let a point D on BC = (3Î» - 2, 1, 4Î»)

Hence,

Area of triangle

(1) 17/âˆš5

(2) 63âˆš5

(3) 205âˆš5

(4) 11/âˆš5

Ans.

Let the plane be

P = (2x + 3y + z + 5) + Î»(x: +y + z- 6) = 0

âˆµ above plane is perpendicular to xy plane.

Hence, the equation of the plane is,

P = x + 2y+ 11 = 0

Distance of the plane P from (0, 0, 256)

(1) 2âˆš13

(2) âˆš91/4

(3) âˆš91/2

(4) âˆš65/2

Ans.

Image of Q (0, -1, -3) in plane is,

(1)

(2) 1/âˆš15

(3)

(4) 1/âˆš30

Ans.

G is the centroid of Î”ABC.

(1) 1

(2) 2

(3) -1

(4) -2

Ans.

Given, (let) and point P (Î², 0, Î²)

Any point on line A = (p, 1, -p -1)

Now, DR of AP â‰¡ <p - Î², 1 - 0 - p - 1 - Î²>

Which is perpendicular to line.

Given that distance AP

âˆ´ Î² = -1

(1) (1, 0, 2)

(2) (2, 0, 1)

(3) (-1, 0, 4)

(4) (4, 0, -1)

Ans.

Let co-ordinates of Q be (Î±, Î², Î³), then

Î± + Î² + Î³ = 3 ...(i)

Î± - Î² + Î³ = 3 ...(ii)

â‡’ Î± + Î³ = 3 and Î² = 0

Equating direction ratio's of PQ, we get

Substituting the values of a and y in equation (i), we get

â‡’ 5Î» + 3 = 3 â‡’ Î» = 0

Hence, point is Q (2, 0, 1)

(1) 9

(2) 15

(3) 5

(4) 13

Ans.

Let,

Given, distance between P

And distance between P

(1) 14

(2) âˆš14

(3) 2âˆš7

(4) 2âˆš14

Ans.

Let points P (3Î» + 2, 2Î» - 1, - Î» + 1) and Q(3Î¼ + 2, 2Î¼ - 1, - Î¼ + 1)

Hence, P (- 1, -3, 2)

Similarly, Q lies on 3x + y + 4z = 16

Hence, Q is (5, 1, 0)

(1) (-1, -4, 1)

(2) (1, 4, -1)

(3) (2, 4, 1)

(4) (2, -4, 1)

Ans.

The equations of angle bisectors are,

â‡’ x - 3y - 2 = 0

or 3x + y + 4z - 6 = 0

(2, -4, 1) lies on the second plane.

(1) 3

(2) 1/3

(3) âˆš3

(4) 1/âˆš3

Ans.

The equation of plane containing two given lines is,

On expanding, we get x - y - z = 0

Now, the length of perpendicular from (2, 1, 4) to this plane

(1) 336

(2) 315

(3) 256

(4) 84

Ans.

(1)

(2)

(3)

(4)

Ans.

Vectors along the given lines L

and

Putting y = 0 in 1

Point on the plane is (â€“5, 0, 4) and normal vector of required plane is

Hence, equation of plane is - 7x + 7y - 8z - 3 = 0

Perpendicular distance is

(1) 2/âˆš3

(2) 2/3

(3) 1/3

(4)

Ans.

D.Râ€™s of AB = (1, 0, 1)

D.Râ€™s of normal to plane = (1, 1, 1)

AB = âˆš2

Length of projection =

(1) 1/8

(2) 25/8

(3) 2

(4) 5

Ans.

(1) 6âˆš5

(2) 3âˆš5

(3) 2âˆš42

(4) âˆš42

Ans.

Equation of PQ,

Let M be (Î»+1, 4Î»- 2, 5Î»+ 3)

As it lies on 2x + 3y â€“ 4z + 22 = 0

Î» = 1

For Q, Î» = 2

Distance PQ

=

(1) 10/âˆš74

(2) 20/âˆš74

(3) 10/âˆš83

(4) 5/âˆš83

Ans.

Let the plane be

a(x - 1)+ b( y + 1) + c(z +1) = 0

It is perpendicular to the given lines

a â€“ 2b + 3c = 0

Solving, a : b : c = 5 : 7 : 3

âˆ´ The plane is 5x + 7y + 3z + 5 = 0

Distance of (1, 3, â€“7) from this plane = 10/âˆš83

(1) (2, -4, 2)

(2) (1, 1, 1)

(3) (0, 0, 0)

(4) (-1, 2, -`= 1)

Ans.

(1)

(2)

(3)

(4)

Ans.

13 x = 6

x = 6/13y

y = 5/13

.... is

(1) 20

(2) 65

(3) 52

(4) 26

Ans.

(1) 1

(2) 2

(3) 3

(4) 0

Ans.

pt (3, -2, Î») on pline 2x - 4y +3z - 2 = 0

= 6 + 8 - 3Î» - 2 = 0

Now

...(1)

...(2)

p (1, 0, 0)

gives are ditersech - thortest distance = 0

x + 8y + 7z = 0

9x + 2y +3z = 0

y + y + z = 0

such that the point (a, b, c) lies on the plane x + 2y + z = 6, then 2a + b + c equals: (2017)

(1) 2

(2) -1

(3) 1

(4) 0

Ans.

(1)

(2)

(3)

(4)

Ans.

(1)

(2)

(3)

(4)

Ans.

Let Centroid be (h, k, l)

âˆ´ x - intp = 3h Y - intp = 3k, 3 - int = 3l

(1) 3âˆš10

(2) 10âˆš3

(3) 10/âˆš3

(4) 20/3

Ans.

Let Q(1, -5, 9)

Line is (say)

Any pt on line we can take P(r + 1, r - 5, r + 9)

So, Pt satisfy Plane

=>(r + 1) - (r - 5) + (r + 9) = 5 r = -10 So, Point P = (-9, -15, -1)

Distance is PQ =

(1) 26

(2) 18

(3) 5

(4) 24

Point on line is P = (3, -2, -4)

'Pâ€™ lies on lx + my - z = 9

=> 3l - 2m + 4 = 9

3l - 2m = 5 ....(1)

As line lies on plane

=> 2 x l + m x (-1) + 3 x (-1) = 0

2l - m = 3 .....(3)

Solving l = 1, m = -1

So, l

(1) 3Ï€/4

(2) Ï€/2

(3) 2Ï€/3

(4) 5Ï€/6

Ans.

Equate,

as unit vectors

(1) (2, 3]

(2) [0, 1)

(3) (3, 4]

(4) [1, 2)

Ans.

(1) 1/âˆš2

(2) 2

(3) âˆš2

(4) 2âˆš2

Ans.

Equation of plane âŠ¥ to the planes.

x - y + 2z = 3 & 2x - 2y + z + 12 = 0

and passes through (1, 2, 2) is

3(x - 1) + 3(y - 2) = 0

x + y = 3 ..... (1)

distance of plane x + y - 3 = 0 from (1, - 2, 4) is

(1) parallel to y-axis

(2) making an acute angle with the positive direction of x-axis

(3) parallel to x-axis

(4) making an obtuse angle with the position direction of x-axis.

Ans.

â‡’ - 8 + 2(9 - 1) - 3 (p + 1) = 0 â‡’ - 3p + 2q - 13 = 0

â‡’ (p, q) lies on line

3x - 2y + 13 = 0

slope = 3/2

(1) 676

(2) 1130

(3) 1348

(4) 1077

Ans.

direction cosine of AD =

â‡’

(1)

(2)

(3)

(4)

Ans.

Position vector of the centroid of

Now we known that centroid divides the line joining orthocentre to circum centre divided by centriod divided by centroid in the ratio in 2 : 1

(1) 3

(2) 2

(3) 1

(4) 4

Ans.

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