Here we provide RD Sharma Class 12 Ex 26.1 Solutions Chapter 26 Scalar Triple Product for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 26.1 Solutions Chapter 26 Scalar Triple Product book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 26 |

Exercise | 26.1 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 26.1 Solutions Chapter 26 Scalar Triple Product**

**Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed **

### Question 1(i). Evaluate the following

**Solution:**

=

=

= 1 + 1 + 1

= 3

### Question 1(ii). Evaluate the following

**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 is 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 is 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 λ so that the following vector is coplanar:

**Solution:**

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

=

= 1(λ -1) + 1(2λ + λ) + 1(-2 – λ)

= λ – 1 + 3λ – 2 -λ

3 = 3λ

1 = λ

So, the value of λ is 1

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

**Solution:**

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

=

= 2(10 + 3 λ) + 1(5 + 3 λ) + 1(λ – 2 λ)

= 20 + 6 λ + 5 + 3 λ – λ

-25 = 8 λ

λ = – 25 / 8

So, the value of λ is -25/8

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

**Solution:**

**Given:**

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

=

= 1(2λ – 2) – 2(6 – 1) – 3(6 – λ)

= 2λ – 2 -12 + 2 -18 + 3λ

= 5λ – 30

30 = 5λ

λ = 6

So, the value of the λ is 6

### Question 5(iv). Find the value of λ 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 – 5λ) + 0

= 5 + 15λ

-5 = 15λ

λ = – 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) ≠ 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 ≠ 0

So, the points are not coplanar.

### Question 9. Find the value of λ 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) + (λ + 1)(15 + 40) = 0

100 – 210 + 55 + 55λ = 0

55λ = 55

λ = 1

So, when the value of λ = 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 c_{1} = 1 and c_{2} = 2, find c_{3} which makes coplanar.

**Solution:**

**Given:**

are coplanar only if = 0

0 – 1(C_{3}) + 1(2) = 0

C_{3} = 2

So, when the value C_{3} = 2, then these points are coplanar.

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

**Solution:**

**Given:**

are coplanar only if = 0

So,

0 – 1 + 1 (C_{1}) = 0

C_{1} = 1

Hence, prove that no value of C_{1 }can make these points coplanar

### Question 13. Find λ for which the points A (3, 2, 1), B (4, λ, 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 =

AD =

On simplifying, we get

1(9) – (λ – 2)(-2 + 9) + 4(3 – 0) = 0

9 – 7 λ + 14 + 12 = 0

7 λ = 35

λ = 5

Hence, the value of λ 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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