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

  • Last Updated : 18 Mar, 2021

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 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)

= -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 is 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) 




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