Class 12 RD Sharma Solutions Chapter 26 Scalar Triple Product Exercise 26.1 (original) (raw)

Last Updated : 27 Aug, 2024

Question 1(i). Evaluate the following [ \hat{i} \hat{j} \hat{k} ] + [ \hat{j} \hat{k} \hat{i} ] + [ \hat{k} \hat{i} \hat{j} ]

**Solution:

[ \hat{i} \hat{j} \hat{k} ] + [ \hat{j} \hat{k} \hat{i} ] + [ \hat{k} \hat{i} \hat{j} ] = ( \hat{i} * \hat{j} ) . \hat{k} + ( \hat{j} * \hat{k} ) . \hat{i} + ( \hat{k} * \hat{i} ) . \hat{j}

= \hat{k} . \hat{k} + \hat{i} . \hat{i} + \hat{j} . \hat{j}

= 1 + 1 + 1

= 3

Question 1(ii). Evaluate the following [ 2\hat{i}\ \hat{j}\ \hat{k} ] + [ \hat{i}\ \hat{k}\ \hat{2i} ] + [ \hat{k} \ \hat{j} \ 2\hat{i} ]

**Solution:

[ 2\hat{i}\ \hat{j}\ \hat{k} ] + [ \hat{i}\ \hat{k}\ \hat{2i} ] + [ \hat{k} \ \hat{j} \ 2\hat{i} ] = ( \hat{2i} * \hat{j} ) . \hat{k} + ( \hat{i} * \hat{k} ) . \hat{j} + ( \hat{k} * \hat{j} ) .2\hat{i}

= 2\hat{k}.\hat{k} + ( -\hat{j}).\hat{j} + ( -\hat{i} ).2\hat{i}

= 2 - 1 - 2

= -1

Question 2(i). Find [ \bar{a}\ \bar{b}\ \bar{c} ] , when \bar{a} = 2\hat{i} - 3\hat{j}, \bar{b} = \hat{i} + \hat{j} - \hat{k} \ and \ \bar{c} = 3\hat{i} - \hat{k}

**Solution:

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 2 & -3 & 0 \\ 1 & 1 & -1 \\ 3 & 0 & -1 \\ \end{vmatrix}= 0

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

= -2 + 6

= 4

Question 2(ii). Find [ \bar{a}\ \bar{b}\ \bar{c} ] , when \bar{a} = \hat{i} - 2\hat{j} + 3\hat{k} , \bar{b} = 2\hat{i} + \hat{j} - \hat{k} \ and\ \bar{c} = \hat{j} + \hat{k}

**Solution:

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 1 & -2 & 3 \\ 2 & 1 &-1 \\ 0 & 1 & 1 \\ \end{vmatrix} = 0

= 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 \bar{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} , \bar{b} = \hat{i} + 2\hat{j} - \hat{k} , \bar{c} = 3\hat{i} - \hat{j} + 2\hat{k}

**Solution:

Volume of a parallelepiped whose adjacent edges are \bar{a},\ \bar{b} ,\ \bar{c} is equal to [ \bar{a}\ \bar{b}\ \bar{c} ]

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 2 & 3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \\ \end{vmatrix} = 0

= 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 \bar{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} , \bar{b} = \hat{i} + 2\hat{j} - \hat{k} , \bar{c} = 3\hat{i} - \hat{j} - 2\hat{k}

**Solution:

Volume of a parallelepiped whose adjacent edges \bar{a},\ \bar{b} ,\ \bar{c} are equal to [ \bar{a}\ \bar{b}\ \bar{c} ]

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1& -2\end{vmatrix} = 0

= 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 \bar{a} = 11\hat{i}, \bar{b} = 2\hat{j} , \bar{c} = 13\hat{k}

**Solution:

Let a = 11\hat{i} , b = 2\hat{j} , c = 13\hat{k}

Volume of a parallelepiped whose adjacent edges are \bar{a},\ \bar{b} ,\ \bar{c} is equal to [ \bar{a}\ \bar{b}\ \bar{c} ]

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 11 & 0& 0 \\ 0 & 2& 0 \\ 0 & 0& 13 \end{vmatrix} = 0

= 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 \bar{a} = \hat{i} + \hat{j} + \hat{k} , \bar{b} = \hat{i} - \hat{j} + \hat{k} , \bar{c} = \hat{i} + 2\hat{j} - \hat{k}

**Solution:

Let \bar{a} = \hat{i} + \hat{j} + \hat{k} , \bar{b} = \hat{i} - \hat{j} + \hat{k} , \bar{c} = \hat{i} + 2\hat{j} - \hat{k}

Volume of a parallelepiped whose adjacent edges \bar{a},\ \bar{b} ,\ \bar{c} are equal to [ \bar{a}\ \bar{b}\ \bar{c} ]

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 1 & 1& 1 \\ 1 & -1& 1 \\ 1 & 2& -1 \end{vmatrix} = 0

= 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 : \bar{a} = \hat{i} + 2\hat{j} - \hat{k}, \bar{b} = 3\hat{i} + 2\hat{j} + 7\hat{k}, \bar{c} = 5\hat{i} + 6\hat{j} + 5\hat{k}

**Solution:

As we know that three vectors \bar{a},\ \bar{b} ,\ \bar{c} are coplanar if their [ \bar{a}\ \bar{b}\ \bar{c} ] = 0.

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 1 & 2 & -1 \\ 3 & 2 & 7 \\ 5 & 6 & 5 \\ \end{vmatrix} = 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 : \bar{a} = -4\hat{i} - 6\hat{j} - 2\hat{k} , \bar{b} = -\hat{i} + 4\hat{j} + 3\hat{k} , \bar{c} = -8\hat{i} - \hat{j} + 3\hat{k}

**Solution:

As we know that three vectors \bar{a},\ \bar{b} ,\ \bar{c} are coplanar if their [ \bar{a}\ \bar{b}\ \bar{c} ] = 0.

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \\ \end{vmatrix} = 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 : \bar{a} = \hat{i} - 2\hat{j} + 3\hat{k} , \bar{b} = -2\hat{i} + 3\hat{j} - 4\hat{k} , \bar{c} = \hat{i} - 3\hat{j} + 5\hat{k}

**Solution:

As we know that three vectors \bar{a},\ \bar{b} ,\ \bar{c} are coplanar if their [ \bar{a}\ \bar{b}\ \bar{c} ] = 0.

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \\ \end{vmatrix}=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: \bar{a} = \hat{i} - \hat{j} + \hat{k} , \bar{b} = 2\hat{i} + \hat{j} - \hat{k} , \bar{c} = λ\hat{i} - \hat{j} + λ\hat{k}

**Solution:

As we know that three vectors \bar{a},\ \bar{b} ,\ \bar{c} are coplanar if their [ \bar{a}\ \bar{b}\ \bar{c} ] = 0.

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 1 &-1 \\ λ & -1 & λ \\ \end{vmatrix} = 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: \bar{a} = 2\hat{i} - \hat{j} + \hat{k} , \bar{b} = \hat{i} + 2\hat{j} - 3\hat{k} , \bar{c} = λ\hat{i} + λ\hat{j} + 5\hat{k}

**Solution:

As we know that three vectors \bar{a},\ \bar{b} ,\ \bar{c} are coplanar if their [ \bar{a}\ \bar{b}\ \bar{c} ] = 0.

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ λ & λ & 5 \\ \end{vmatrix} = 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:\bar{a} = \hat{i} + 2\hat{j} - 3\hat{k} , \bar{b} = 3\hat{i} + λ\hat{j} + \hat{k} , \bar{c} = \hat{i} + 2\hat{j} + 2\hat{k}

**Solution:

**Given:

\bar{a} = \hat{i} + 2\hat{j} - 3\hat{k} \\ \bar{b} = 3\hat{i} + λ\hat{j} + \hat{k} \\ \bar{c} = \hat{i} + 2\hat{j} + 2\hat{k}

As we know that three vectors \bar{a},\ \bar{b} ,\ \bar{c} are coplanar if their [ \bar{a}\ \bar{b}\ \bar{c} ] = 0.

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 1 & 2 & -3 \\ 3 & λ & 1 \\ 1 & 2 & 2 \\ \end{vmatrix} = 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: \bar{a} = \hat{i} + 3\hat{j} , \bar{b} = 5\hat{k} , \bar{c} = λ \bar{i} - \hat{j}

**Solution:

**Given:

\bar{a} = \hat{i} + 3\hat{j} \\ \bar{b} = 5\hat{k} \\ \bar{c} = λ \bar{i} - \hat{j}

So, to prove that these points are coplanar, we have to prove that [ \bar{a}\ \bar{b}\ \bar{c} ] = 0

[ \bar{a}\ \bar{b}\ \bar{c} ] = \begin{vmatrix} 1 & 3 & 0 \\ 0 & 0 & 5 \\ λ & -1 & 0 \\ \end{vmatrix} = 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 \hat{6i} - 7\hat{j} , 16\hat{i} - 19\hat{j} - 4\hat{k} , 3\hat{j} - 6\hat{k}, 2\hat{i} + 5\hat{j} + 10\hat{k} are not coplanar.

**Solution:

Let us considered

OA = 6\hat{i} - 7\hat{j}

OB = 16\hat{i} - 19\hat{j} - 4\hat{k}

OC = 3\hat{j} - 6\hat{k}

OD = 2\hat{i} + 5\hat{j} + 10\hat{k}

AB = OB - OA = 16\hat{i} - 25\hat{j} - 4\hat{k}

AC = OC - OA = -16\hat{i} - 16\hat{j} + 2\hat{k}

CD = OD - OC = 2\hat{i} + 2\hat{j} + 16\hat{k}

AD = OD - OA = 4\hat{i} + 12\hat{j}+ 10\hat{k}

So, to prove that these points are coplanar, we have to prove that [\overline{AB} \ \overline{AC} \ \overline{AD} ] =0

\begin{vmatrix} 16 & -25 & -4 \\ -16& -16& 2 \\ -4 & 12 & 10 \\ \end{vmatrix}

= 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)

\overline{AB} = 4\hat{i} - 2\hat{j} - 2\hat{k}

\overline{AC} = -2\hat{i} + 4\hat{j} - 2\hat{k}

\overline{AD} = -2\hat{i} - 2\hat{j} + 4\hat{k}

So, to prove that these points are coplanar, we have to prove that [\overline{AB} \ \overline{AC} \ \overline{AD}] = 0

Thus, \begin{vmatrix} 4 &-2 & -2 \\ -2& 4& -2 \\ -2 & -2& 4 \\ \end{vmatrix}

= 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 6\hat{i} - 7\hat{j} , 16\hat{i} - 19\hat{j} - 4\hat{k} , 3\hat{i} - 6\hat{k}, 2\hat{i} - 5\hat{j} + 10\hat{k}

**Solution:

Let us considered

OA = 6\hat{i} - 7\hat{j}

OB = 16\hat{i} - 19\hat{j} - 4\hat{k}

OC = 3\hat{i} - 6\hat{k}

OD = 2\hat{i} -5\hat{j} + 10\hat{k}

Thus,

AB = OB - OA = 10\hat{i} - 12\hat{j} - 4\hat{k}

AC = OC - OA = -3\hat{i} + 7\hat{j}- 6\hat{k}

AD = OD - OA = -4\hat{i} + 2\hat{j} + 10\hat{k}

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

On simplifying, we get

\begin{vmatrix} 10 & -12 & -4 \\ -3 & 7 & -6\\ -4 & 2 & 10 \\ \end{vmatrix}

= 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 -\hat{j} - \hat{k}, 4\hat{i} + 5\hat{j} + λ \hat{k}, 3\hat{i} + 9\hat{j} + 4\hat{k}\ and\ -4\hat{i} + 4\hat{j} + 4\hat{k} are coplanar

**Solution:

Let us considered:

Position vector of A = -\hat{j} -\hat{k}

Position vector of B = 4\hat{i} + 5\hat{j} + λ\hat{k}

Position vector of C = 3\hat{i} + 9\hat{j} + 4\hat{k}

Position vector of D = -4\hat{i} + 4\hat{j} + 4\hat{k}

If the given vectors \overline{AB},\ \overline{AC}, \ \overline{AD} are coplanar, then the four points are coplanar

\overline{AB} = 4\hat{i} + 6\hat{j} + ( λ + 1 ) \hat{k}

\overline{AC} = 3\hat{i} + 10\hat{j} + 5\hat{k}

\overline{AD} = -4\hat{i} + 5\hat{j} + 5\hat{k}

On simplifying, we get

\begin{vmatrix} 4 & 6 & λ+1 \\ 3 & 10 & 5 \\ -4 & 5 & 5 \end{vmatrix} = 0

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 ( \bar{a} - \bar{b} ) . [( \bar{b} - \bar{c} ) * (\bar{ c} - \bar{a} ) ] = 0

**Solution:

**Given: ( \bar{a} - \bar{b} ) . [( \bar{b} - \bar{c} ) * (\bar{ c} - \bar{a} ) ] = 0

One solving the given equation we get

= [ ( \bar{a} - \bar{b} ) ( \bar{b} - \bar{c} ) ( \bar{c} - \bar{a} ) ]

= [ a( \bar{b} - \bar{c} ) ( \bar{c} - \bar{a} ) ] + [ -b ( \bar{b} - \bar{c} ) ( \bar{c} - \bar{a} ) ]

= 6 [ a b c ] - 6 [ a b c ]

= 0

Hence proved

Question 11. \bar{a} , \bar{b} ,\bar{c} are the position vectors of points A, B and C respectively, prove that \bar{a} * \bar{b} + \bar{b} * \bar{c} + \bar{c} * \bar{a} is a vector perpendicular to the plane of triangle ABC.

**Solution:

In the given triangle ABC,

If \bar{a} = AB

\bar{b} = BC

\bar{c} = AC

Then,

\bar{a} * \bar{b} is perpendicular to the plane of the given triangle ABC

\bar{b} * \bar{c} is perpendicular to the plane of the given triangle ABC

\bar{c} * \bar{a} is perpendicular to the plane of the given triangle ABC

Hence, proved that \bar{a} * \bar{b} + \bar{b} * \bar{c} + \bar{c} * \bar{a}

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

Question 12(i). Let \bar{a} = \hat{i} + \hat{j} + \hat{k}, \ \bar{b} = \hat{i} \ and \ \bar{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k} . Then, if c1 = 1 and c2 = 2, find c3 which makes \bar{a}, \bar{b}, \bar{c} coplanar.

**Solution:

**Given:

\bar{a} = \hat{i} + \hat{j} + \hat{k}, \\ \bar{b} = \hat{i} \\ \bar{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}

\bar{a}, \bar{b}, \bar{c} are coplanar only if [\bar{a}, \bar{b}, \bar{c}] = 0

\begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & C_3 \end{vmatrix} = 0

0 - 1(C3) + 1(2) = 0

C3 = 2

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

Question 12(ii). Let\bar{a} = \hat{i}+\hat{j}+\hat{k}, \bar{b} = \hat{i} and \bar{c} = c1\hat{i} + c2\hat{j} + c3\hat{k} . Then, if c2 = -1 and c3 =1, show that no value of c1 can make \bar{a}, \bar{b}, \bar{c} coplanar

**Solution:

**Given:

\bar{a} = \hat{i}+\hat{j}+\hat{k}\\ \bar{b} = \hat{i}\\ \bar{c} = c1\hat{i} + c2\hat{j} + c3\hat{k}

\bar{a}, \bar{b}, \bar{c} are coplanar only if [\bar{a}, \bar{b}, \bar{c}] = 0

So,

\begin{vmatrix} 1 & 1 & 1\\ 1 & 0 & 0\\ -1& C_1& 1\end{vmatrix}

0 - 1 + 1 (C1) = 0

C1 = 1

Hence, prove that no value of C1 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 = 3\hat{i} + 2\hat{j}+ \hat{k}

Position vector of OB = 4\hat{i} + λ\hat{j} + 5\hat{k}

Position vector of OC = 4\hat{i} + 2\hat{j} - 2\hat{k}

Position vector of OD = 6\hat{i} + 5\hat{j} - \hat{k}

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

AB = \hat{i} + ( λ - 2 ) \hat{j} + 4\hat{k}

AC = \hat{i} + 0\hat{j} - 3\hat{k}

AD = 3\hat{i} + 3\hat{j} - 2\hat{k}

On simplifying, we get

\begin{vmatrix} 1 & ( λ - 2 ) & 4 \\ 1 & 0 & -3 \\ 3 & 3 & -2 \\ \end{vmatrix}

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)

Summary

The scalar triple product of three vectors a, b, and c is defined as the dot product of one vector with the cross product of the other two. It's denoted as [a b c] or a · (b × c). Key points include:

1. It represents the volume of the parallelepiped formed by the three vectors.

2. It's invariant under cyclic permutation of vectors.

3. Its value is zero if the vectors are coplanar.

4. It can be calculated using the determinant method.

Practice Problems

**1. Calculate the scalar triple product of a = 2i + 3j - k, b = i - 2j + 4k, and c = 3i + j + 2k.

**2. Prove that the vectors a = i + 2j - k, b = 2i - j + 3k, and c = i + 4j + 2k are coplanar.

**3. Find the volume of the parallelepiped formed by the vectors a = 3i - j + 2k, b = i + 2j - k, and c = 2i + j + 3k.

**4. If a = 2i - j + k, b = i + 3j - 2k, and c = 3i - 2j + 4k, find the value of [a b c] + [b c a] + [c a b].

**5. Prove that [a b c] = [b c a] = [c a b].

**6. If a · (b × c) = 6, b · (c × a) = -6, and c · (a × b) = 6, find the value of (a × b) · c.

**7. Show that [a + b, b + c, c + a] = 2[a b c].

**8. If a, b, and c are unit vectors such that a + b + c = 0, prove that [a b c] = ±√3/2.

**9. Prove that [a b (a × b)] = |a × b|².

**10. If a = 3i + 2j - k, b = i - j + 2k, and c = 2i + 3j + k, find λ such that [a b (c + λa)] = 0.