# Class 12 RD Sharma Solutions – Chapter 26 Scalar Triple Product – Exercise 26.1

Solution:

=

= 1 + 1 + 1

= 3

Solution:

=

=

= 2 – 1 – 2

= -1

### Question 2(i). Find , when

Solution:

=

= 2(-1 – 0) + 3(-1 + 3)

= -2 + 6

= 4

### Question 2(ii). Find, when

Solution:

= 1(1 + 1) + 2(2 + 0) + 3(2 – 0)

= 2 + 4 + 6

= 12

### Question 3(i). Find the volume of the parallelepiped whose coterminous edges are represented by vector

Solution:

Volume of a parallelepiped whose adjacent edges are  is equal to

=

= 2(4 – 1) – 3(2 + 3) + 4(-1 – 6)

= 6 – 15 – 28

= -9 – 28

= -37

So, Volume of parallelepiped is | -37 | = 37 cubic unit.

### Question 3(ii). Find the volume of the parallelepiped whose coterminous edges are represented by vector

Solution:

Volume of a parallelepiped whose adjacent edges  are  equal to

=

= 2(-4 – 1) + 3(-2 + 3) + 4(-1 – 6)

= -10 + 3 – 28

= -10 – 25

= -35

So, Volume of parallelepiped = | -35 | = 35 cubic unit.

### Question 3(iii). Find the volume of the parallelepiped whose coterminous edges are represented by vector

Solution:

Let a = 11, b = 2, c = 13

Volume of a parallelepiped whose adjacent edges are  is equal to

=

= 11(26 – 0) + 0 + 0

= 286

Volume of a parallelepiped = | 286| = 286 cubic units.

### Question 3(iv). Find the volume of the parallelepiped whose coterminous edges are represented by vector

Solution:

Let

Volume of a parallelepiped whose adjacent edges  are  equal to

=

= 1(1 – 2) – 1(-1 – 1) + 1(2 + 1)

= -1 + 2 + 3

= 4

Volume of a parallelepiped = |4| = 4 cubic units.

### Question 4(i). Show of the following triads of vector is coplanar :

Solution:

As we know that three vectors  are coplanar if their  = 0.

=

= 1(10 – 42) – 2(15 – 35) – 1(18 – 10)

= -32 + 40 – 8

= 0

So, the given vectors are coplanar.

### Question 4(ii). Show of the following triads  of vector is coplanar :

Solution:

As we know that three vectors  are coplanar if their  = 0.

=

= -4(12 + 3) + 6(-3 + 24) – 2(1 + 32)

= -60 + 126 – 66

= 0

So, the given vectors are coplanar.

### Question 4(iii). Show of the following triads of vector is coplanar :

Solution:

As we know that three vectors  are coplanar if their  = 0.

=

= 1(15 – 12) + 2(-10 + 4) + 3(6 – 3)

= 3 – 12 + 9

= 0

So, the given vectors are coplanar.

### Question 5(i). Find the value of  Î» so that the following vector is coplanar:

Solution:

As we know that three vectors  are coplanar if their  = 0.

=

= 1(Î» -1) + 1(2Î» + Î») + 1(-2 – Î»)

= Î» – 1 + 3Î» – 2 -Î»

3 = 3Î»

1 = Î»

So, the value of Î» is 1

### Question 5(ii). Find the value of  Î» so that the following vector is coplanar:

Solution:

As we know that three vectors  are coplanar if their = 0.

=

= 2(10 + 3 Î») + 1(5 + 3 Î») + 1(Î»  – 2 Î»)

= 20 + 6 Î» + 5 + 3 Î» – Î»

-25 = 8 Î»

Î»  = – 25 / 8

So, the value of Î» is -25/8

### Question 5(iii). Find the value of  Î» so that the following vector is coplanar:

Solution:

Given:

As we know that three vectors  are coplanar if their  = 0.

=

= 1(2Î» – 2) – 2(6 – 1) – 3(6 – Î»)

= 2Î» – 2 -12 + 2 -18 + 3Î»

= 5Î» – 30

30 = 5Î»

Î» = 6

So, the value of the Î» is 6

### Question 5(iv). Find the value of Î» so that the following vector is coplanar:

Solution:

Given:

So, to prove that these points are coplanar, we have to prove that  = 0

=

= 1(0 + 5) – 3(0 – 5Î») + 0

= 5 + 15Î»

-5 = 15Î»

Î» = – 1 / 3

### Question 6. Show that the four points having position vectors  are not coplanar.

Solution:

Let us considered

OA =

OB =

OC =

OD =

AB = OB – OA =

AC = OC – OA =

CD = OD – OC =

AD = OD – OA =

So, to prove that these points are coplanar, we have to prove that

= 16(-160 – 24) + 25(-160 + 8) – 4(-144 + 64) â‰  0

Hence, proved that the points are not coplanar.

### Question 7. Show that the points A (-1, 4, -3), B(3, 2, -5), C(-3, 8, -5), and D(-3, 2, 1) are coplanar

Solution:

Given:

A = (-1, 4, -3)

B = (3, 2, -5)

C = (-3, 8, -5)

D = (-3, 2, 1)

=

=

=

So, to prove that these points are coplanar, we have to prove that

Thus,

= 4[16 – 4] + 2[-8 -4] – 2[4 + 8]

= 48 – 24 – 24 = 0

Hence, proved.

### Question 8. Show that four points whose position vectors are

Solution:

Let us considered

OA =

OB =

OC =

OD =

Thus,

AB = OB – OA =

AC = OC – OA =

AD = OD – OA =

If the vectors AB, AC and AD are coplanar then the four points are coplanar

On simplifying, we get

= 10(70 + 12) + 12(-30 – 24) – 4(-6 + 28)

= 820 – 648 – 88

= 84 â‰  0

So, the points are not coplanar.

### Question 9. Find the value of  Î» for which the four points with position vectors  are coplanar

Solution:

Let us considered:

Position vector of A =

Position vector of B =

Position vector of C =

Position vector of D =

If the given vectors  are coplanar, then the four points are coplanar

=

=

=

On simplifying, we get

4(50 – 25) – 6(15 + 20) + (Î» + 1)(15 + 40) = 0

100 – 210 + 55 + 55Î» = 0

55Î» = 55

Î» = 1

So, when the value of Î» = 1, the given points are coplanar.

### Question 10. Prove that

Solution:

Given:

One solving the given equation we get

=

= 6 [ a  b  c ] – 6 [ a  b  c ]

= 0

Hence proved

### Question 11. are the position vectors of points A, B and C respectively, prove that  is a vector perpendicular to the plane of triangle ABC.

Solution:

In the given triangle ABC,

If  = AB

= BC

= AC

Then,

is perpendicular to the plane of the given triangle ABC

is perpendicular to the plane of the given triangle ABC

is perpendicular to the plane of the given triangle ABC

Hence, proved that

is a vector perpendicular to the plane of the given triangle ABC.

### Question 12(i). Let . Then, if c1 = 1 and c2 = 2, find c3 which makes  coplanar.

Solution:

Given:

are coplanar only if  = 0

0 – 1(C3) + 1(2) = 0

C3 = 2

So, when the value C3 = 2, then these points are coplanar.

### Question 12(ii). Let and . Then, if c2 = -1 and c3 =1, show that no value of c1 can make  coplanar

Solution:

Given:

are coplanar only if = 0

So,

0 – 1 + 1 (C1) = 0

C1 = 1

Hence, prove that no value of C1 can make these points coplanar

### Question 13. Find Î» for which the points A (3, 2, 1), B (4,  Î», 5), C (4, 2, -2), and D (6, 5, -1) are coplanar

Solution:

Let us considered:

Position vector of OA =

Position vector of OB =

Position vector of OC =

Position vector of OD =

If the vectors AB, AC, and AD are coplanar, then the four points are coplanar

AB =

AC =

On simplifying, we get

1(9) – (Î» – 2)(-2 + 9) + 4(3 – 0) = 0

9 – 7 Î» + 14 + 12 = 0

7 Î» = 35

Î» = 5

Hence, the value of Î» is 5. So the coplanar points are, A(3, 2, 1), B(4, 5, 5), C(4, 2, -2), and D(6, 5, -1)

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