RD Sharma Class 12 Exercise 25 MCQ Scalar Triple Product Solutions Maths

RD Sharma Class 12 Exercise 25 MCQ Scalar Triple Product Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 24, 2022 09:16 PM IST

The syllabus in the class 12 books is tricky and challenging for the students. Many students face hardships when they try to work out the mathematics homework without any help. Even though a teacher explains to them the concept and the steps in which a sum should be solved, not every student can grasp in an instance. The RD Sharma Class 12th Chapter 25 MCQ lends a helping hand to the students who desperately need proper guidance.

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RD Sharma Class 12 Solutions Chapter25MCQ Scalar Triple Product - Other Exercise

Scalar Triple Product Excercise: 25 MCQ

Scalar triple product exercise multiple choice question 1

Answer:
0
Hint:
Use formula, that when vectors are co planar, they are, scalar triple product is zero.
Given:
$\overrightarrow{a}$ lies on the plane of $\overrightarrow{b}$ and $\overrightarrow{c}$ . If
$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ lies on one plane they are coplanar.
$\therefore \left [\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \right ]=0$

Scalar triple product exercise multiple choice question 2

Answer:
0
Hint:
Use formula of scalar triple product
Given:
$\left | \overrightarrow{a} \right |=1,\: \left |\overrightarrow{b} \right |=5,\: \left | \overrightarrow{c} \right |=3$
Now,
$\begin{aligned} &\left [ \overrightarrow{a}-\overrightarrow{b} \qquad \overrightarrow{b}-\overrightarrow{c} \qquad \overrightarrow{c} -\overrightarrow{a} \right ] \\ &=\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]-\left [ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right ] \end{aligned}$
(From scalar product)
$\begin{aligned} &=\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]+\left [ \overrightarrow{a}\overrightarrow{c}\overrightarrow{b} \right ]\\ &=\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]-\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]\\ &=0 \end{aligned}$

Scalar triple product exercise multiple choice question 3

Answer:
$\pm 1$
Hint:
Use formula of scalar triple product
Given:
$\overrightarrow{a},\overrightarrow{b}\: and\; \overrightarrow{c}$ be the non-coplanar and mutually perpendicular unit vectors.
$\therefore \left |\overrightarrow{a} \right |=\left |\overrightarrow{b} \right |= \left |\overrightarrow{c} \right |=1$
Now
$\begin{aligned} &\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]=\left | \overrightarrow{a}\times \overrightarrow{b} \right |.\overrightarrow{c}\\ &=\left | \overrightarrow{a}\times \overrightarrow{b} \right |.1\\ &\because \overrightarrow{c} \text { is unit vector }\\ &\therefore \left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]=\left | \overrightarrow{a}\times \overrightarrow{b} \right | \text { or } -\left | \overrightarrow{a}\times \overrightarrow{b} \right |\\ &\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |sin \: \theta \dots \dots \dots (1)\\ &-\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |sin \: \theta \dots \dots \dots (2)\\ &=(1)(1)sin\frac{\pi }{2}\\ &=1 \dots \dots \dots From(1)\\ &=-(1)(1)sin\frac{\pi }{2}\\ &=-1 \dots \dots \dots From(2) \end{aligned}$
Thus value could be
$\begin{aligned} &\pm 1 \end{aligned}$

Scalar triple product exercise multiple choice question 4

Answer:
0
Hint:
When dot product is zero either both are perpendicular or one of the vectors is zero vector.
Given:
$\overrightarrow{r}.\overrightarrow{a}=\overrightarrow{r}.\overrightarrow{b}=\overrightarrow{r}.\overrightarrow{c}=0$
If
$\overrightarrow{r}.\overrightarrow{a}=0$
either
$\overrightarrow{a}=0$
or both
$\overrightarrow{r}$ and $\overrightarrow{a}$ are perpendicular to each other.
Thus, for
$\left [\overrightarrow{a} \overrightarrow{b}\overrightarrow{c} \right ]$
when all three are non-zero. Scalar triple product is zero if they are coplanar & perpendicular to
$\overrightarrow{r}$
Thus
$\left [\overrightarrow{a} \overrightarrow{b}\overrightarrow{c} \right ]=0$

Scalar triple product exercise multiple choice question 5

Answer:
0
Hint:
Simplify the expression
Given:
$\begin{aligned} &(\overrightarrow{a}-\overrightarrow{b}).\left \{ (\overrightarrow{b}-\overrightarrow{c})\times (\overrightarrow{c}-\overrightarrow{a}) \right \}\\ &=(\overrightarrow{a}-\overrightarrow{b}).\left \{ \overrightarrow{b}\times (\overrightarrow{c}-\overrightarrow{a})-\overrightarrow{c}\times (\overrightarrow{c}-\overrightarrow{a}) \right \}\\ &=(\overrightarrow{a}-\overrightarrow{b}).\left \{ (\overrightarrow{b}\times \overrightarrow{c})-(\overrightarrow{b}\times \overrightarrow{a})-(\overrightarrow{c}\times \overrightarrow{c})+(\overrightarrow{c}\times \overrightarrow{a}) \right \} \end{aligned}$
$\begin{aligned} &=(\overrightarrow{a}-\overrightarrow{b}).\left \{ (\overrightarrow{b}\times \overrightarrow{c})-(\overrightarrow{b}\times \overrightarrow{a})+(\overrightarrow{c}\times \overrightarrow{a}) \right \} \end{aligned}$
$\begin{aligned} &=\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})-\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{a})+\overrightarrow{a}.(\overrightarrow{c}\times \overrightarrow{a})-\overrightarrow{b}.(\overrightarrow{b}\times \overrightarrow{c})+\overrightarrow{b}.(\overrightarrow{b}\times \overrightarrow{a})-\overrightarrow{b}.(\overrightarrow{c}\times \overrightarrow{a})\\ &=\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]-\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{a} \right ]+\left [ \overrightarrow{a}\overrightarrow{c}\overrightarrow{a} \right ]-\left [ \overrightarrow{b}\overrightarrow{b}\overrightarrow{c} \right ]+\left [ \overrightarrow{b}\overrightarrow{b}\overrightarrow{a} \right ]-\left [ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right ] \end{aligned}$
$\begin{aligned} &=\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]-0+0-0+0-\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]\\ &=0 \end{aligned}$

Scalar triple product exercise multiple choice question 6

Answer:
0
Hint:
Use scalar triple product & Simplify
Given:
$\frac{\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})}{(\overrightarrow{c}\times \overrightarrow{a}).\overrightarrow{b}}+\frac{\overrightarrow{b}.(\overrightarrow{a}\times \overrightarrow{c})}{\overrightarrow{c}.(\overrightarrow{a}\times \overrightarrow{b})}$
We know,
$\begin{aligned} &\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]=\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})\\ &\therefore \frac{\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]}{\left [ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right ]}+\frac{\left [ \overrightarrow{b}\overrightarrow{a}\overrightarrow{c} \right ]}{\left [\overrightarrow{c}\overrightarrow{a}\overrightarrow{b} \right ]}\\ &\left ( \because \left [ \overrightarrow{b}\overrightarrow{c}\overrightarrow{a} \right ]=\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ] \text { and } \left [ \overrightarrow{c}\overrightarrow{a}\overrightarrow{b} \right ]=-\left [ \overrightarrow{b}\overrightarrow{a}\overrightarrow{c} \right ] \right )\\ &=1-1\\ &=0 \end{aligned}$

Scalar triple product exercise multiple choice question 7

Answer:
$=\frac{\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2}{4}$
Hint:
Use scalar triple product &
$sin \frac{\pi }{6}=\frac{1}{2}$
Given:
$\overrightarrow{a},\overrightarrow{b}$ are two vectors &
$\overrightarrow{c}$ is unit perpendicular to both
$\therefore \left |\overrightarrow{c} \right |=1$
For
$\begin{aligned} &\begin{vmatrix} a_1 &a_2 &a_3 \\ b_1 &b_2 &b_3 \\ c_1 &c_2 &c_3 \end{vmatrix}^2\\ &=\left ( \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right )^2\\ &=\left ( \left | \overrightarrow{a}\times \overrightarrow{b} \right |.\overrightarrow{c} \right )^2\\ &=\left | \overrightarrow{a}\times \overrightarrow{b} \right |^2 \end{aligned}$
$\begin{aligned} &\because \overrightarrow{c} \text { is unit vector }\\ &=\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2sin^2\frac{\pi }{6}\\ &=\frac{\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2}{4} \end{aligned}$

Scalar triple product exercise multiple choice question 8

Answer:

Hint: $\text {Volume of parallelopiped} = \left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]$
Where,
$\overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \text { are adjacent edges }$
Given:
Adjacent edges
$\begin{aligned} &\overrightarrow{a}+\overrightarrow{b}, \overrightarrow{b}+\overrightarrow{c},\overrightarrow{c}+\overrightarrow{a} \text { and }\\ &\overrightarrow{a}=2\widehat{i}-3\widehat{j}+5\widehat{k}\\ &\overrightarrow{b}=3\widehat{i}-4\widehat{j}+5\widehat{k}\\ &\overrightarrow{c}=5\widehat{i}-3\widehat{j}-2\widehat{k}\\ &\therefore \overrightarrow{a}+\overrightarrow{b}=(2+3)\overrightarrow{i}+(-3-4)\overrightarrow{j}+(5+5)\overrightarrow{k}\\ &=5\overrightarrow{i}-7\overrightarrow{j}+10\overrightarrow{k} \end{aligned}$
$\begin{aligned} &\overrightarrow{b}+\overrightarrow{c}=(3+5)\widehat{i}+(-4-3)\widehat{j}+(5-2)\widehat{k}\\ &=8\widehat{i}-7\widehat{j}+3\widehat{k}\\ &\overrightarrow{c}+\overrightarrow{a}=(5+2)\widehat{i}+(-3-3)\widehat{j}+(-2+5)\widehat{k}\\ &=7\widehat{i}-6\widehat{j}+3\widehat{k} \end{aligned}$
$\begin{aligned} &\therefore \end{aligned}$ Volume of parallelopiped=
$\begin{aligned} &\begin{vmatrix} 5 &-7 &10 \\ 8 &-7 &3 \\ 7 &-6 &3 \end{vmatrix}\\ &=5(-21+18)+7(24-21)+10(-48+49)\\ &=-15+21+10\\ &=16 \end{aligned}$
Note : Given all option are incorrect

Scalar triple product exercise multiple choice question 9

Answer:
6
Hint:
Simplify the expression
Given:
$\begin{aligned} &\left [ 2\overrightarrow{a}+4\overrightarrow{b} \quad \overrightarrow{c} \quad \overrightarrow{d} \right ]=\lambda \left [ \overrightarrow{a} \quad \overrightarrow{c} \quad \overrightarrow{d} \right ]+\mu \left [ \overrightarrow{b} \quad \overrightarrow{c} \quad \overrightarrow{d} \right ]\\ &\left [ 2\overrightarrow{a}+4\overrightarrow{b} \quad \overrightarrow{c} \quad \overrightarrow{d} \right ]= \left [ 2\overrightarrow{a} \quad \overrightarrow{c} \quad \overrightarrow{d} \right ]+\left [ 4\overrightarrow{b} \quad \overrightarrow{c} \quad \overrightarrow{d} \right ]\\ &=2 \left [ \overrightarrow{a} \quad \overrightarrow{c} \quad \overrightarrow{d} \right ]+4\left [ \overrightarrow{b} \quad \overrightarrow{c} \quad \overrightarrow{d} \right ]\\ \end{aligned}$
Comparing we get
$\begin{aligned} &\lambda =2,\mu =4\\ &\lambda +\mu =2+4\\ &\lambda +\mu =6 \end{aligned}$

Scalar triple product exercise multiple choice question 10

Answer:
$=\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2$
Hint:
Simplify expression
Given:
$\begin{aligned} &\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{a}\times \overrightarrow{b} \right ]+\left [ \overrightarrow{a}. \overrightarrow{b} \right ]^2\\ &=(\overrightarrow{a}\times \overrightarrow{b}).(\overrightarrow{a}\times \overrightarrow{b})+\left [ \overrightarrow{a}. \overrightarrow{b} \right ]^2\\ &=(\overrightarrow{a}\times \overrightarrow{b})^2+\left [ \overrightarrow{a}. \overrightarrow{b} \right ]^2\\ &=\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2sin^2\theta +\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2cos^2\theta\\ &=\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2(sin^2\theta+cos^2\theta )\\ &=\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2 \end{aligned}$

Scalar triple product exercise multiple choice question 11

Answer:
10
Hint:
If vectors are coplanar then
$\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ]=0$
Given:
$\begin{aligned} &\begin{vmatrix} 4 &11 &m \\ 7 &2 &6 \\ 1 &5 &4 \end{vmatrix}=0 \end{aligned}$
$\begin{aligned} &\therefore \end{aligned}$ Solving,
$\begin{aligned} &4(8-30)-11(28-6)+m(35-2)\\ &=-88-242+33m\\ &=-330+33m\\ \end{aligned}$
Now,
$\begin{aligned} &=-330+33m\\ &33m=330\\ &m=\frac{330}{33}\\ &m=10 \end{aligned}$

Scalar triple product exercise multiple choice question 12

Answer:
$\overrightarrow{a}.\overrightarrow{b}=0,\overrightarrow{b}.\overrightarrow{c}=0,\overrightarrow{c}.\overrightarrow{a}=0$
Hint:
Use scalar triple product
Given:
$(\overrightarrow{a}.\overrightarrow{b}).\overrightarrow{c}=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\left | \overrightarrow{c} \right |$
For
$(\overrightarrow{a}.\overrightarrow{b}).\overrightarrow{c}=\left | \overrightarrow{a}\times \overrightarrow{b} \right |\left | \overrightarrow{c} \right |cos\theta$
where,
$\theta$ is angle between cross product of
$(\overrightarrow{a}\times \overrightarrow{b}) \text { and }\overrightarrow{c}$
Similarly,
$(\overrightarrow{a}\times \overrightarrow{b}) .\overrightarrow{c}=\left ( \left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |sin\beta \right ).\overrightarrow{c}$
where
$\beta$ is angle between
$\overrightarrow{a} \text { and } \overrightarrow{b}$
Thus, for all
$\overrightarrow{a}.\overrightarrow{b}=0,\overrightarrow{b}.\overrightarrow{c}=0,\overrightarrow{c}.\overrightarrow{a}=0$

Scalar triple product exercise multiple choice question 13

Answer:
$\begin{aligned} &=\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ] \end{aligned}$
Hint:
Simplify the expression
Given:
$\begin{aligned} &(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{b}+\overrightarrow{c})\times (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\\ &=(\overrightarrow{a}+\overrightarrow{b}).\left [ \overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}+\overrightarrow{c}\times \overrightarrow{b}+\overrightarrow{c}\times \overrightarrow{c} \right ] \end{aligned}$
$\begin{aligned} &=(\overrightarrow{a}+\overrightarrow{b}).\left [ \overrightarrow{b}\times \overrightarrow{a}+0+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}+\overrightarrow{c}\times \overrightarrow{b}+0 \right ]\\ &=(\overrightarrow{a}+\overrightarrow{b}).\left [ (\overrightarrow{b}\times \overrightarrow{a})+(\overrightarrow{c}\times \overrightarrow{a}) \right ] \end{aligned}$
$\begin{aligned} &=\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{a})+\overrightarrow{a}.(\overrightarrow{c}\times \overrightarrow{a})+\overrightarrow{b}.(\overrightarrow{b}\times \overrightarrow{a})+\overrightarrow{b}.(\overrightarrow{c}\times \overrightarrow{a}) \\ &=\overrightarrow{a}\overrightarrow{b}\overrightarrow{a}+\overrightarrow{a}\overrightarrow{c}\overrightarrow{a}+\overrightarrow{b}\overrightarrow{b}\overrightarrow{a}+\overrightarrow{b}\overrightarrow{c}\overrightarrow{a}\\ &=0+0+0+\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\\ &=\left [ \overrightarrow{a}\quad \overrightarrow{b}\quad \overrightarrow{c} \right ] \end{aligned}$

Scalar triple product exercise multiple choice question 14

Answer:
$\begin{aligned} &=-\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ] \end{aligned}$
Hint:
Simplify the expression
Given:
$\begin{aligned} &\left ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \right ).\left [ \left ( (\overrightarrow{a} + \overrightarrow{b})\times (\overrightarrow{a} + \overrightarrow{c}) \right ) \right ]\\ &=\left ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \right ).\left [ (\overrightarrow{a}\times \overrightarrow{a})+(\overrightarrow{a}\times\overrightarrow{c})+(\overrightarrow{b}\times\overrightarrow{a})+(\overrightarrow{b}\times\overrightarrow{c}) \right ]\\ &=\left ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \right ).\left [ 0+(\overrightarrow{a}\times\overrightarrow{c})+(\overrightarrow{b}\times\overrightarrow{a})+(\overrightarrow{b}\times\overrightarrow{c}) \right ]\\ \end{aligned}$
$\begin{aligned} &=\overrightarrow{a}.(\overrightarrow{a}\times \overrightarrow{c})+ \overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{a})+ \overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})+ \overrightarrow{b}.(\overrightarrow{a}\times \overrightarrow{c})+ \overrightarrow{b}.(\overrightarrow{b}\times \overrightarrow{a})+ \overrightarrow{b}.(\overrightarrow{b}\times \overrightarrow{c})+\overrightarrow{c}.(\overrightarrow{a}\times \overrightarrow{c})+ \overrightarrow{c}.(\overrightarrow{b}\times \overrightarrow{a})+ \overrightarrow{c}.(\overrightarrow{b}\times \overrightarrow{c}) \end{aligned}$
$\begin{aligned} &=\left [ \overrightarrow{a} \quad \overrightarrow{a} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{a} \right ]+\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{b} \quad \overrightarrow{a} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{b} \quad \overrightarrow{b} \quad \overrightarrow{a} \right ]+\left [ \overrightarrow{b} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{c} \quad \overrightarrow{a} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{c} \quad \overrightarrow{b} \quad \overrightarrow{a} \right ]+\left [ \overrightarrow{c} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ] \end{aligned}$
$\begin{aligned} &=0+0+\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ]+\left [ \overrightarrow{b} \quad \overrightarrow{a} \quad \overrightarrow{c} \right ]+0+0+0+\left [ \overrightarrow{c} \quad \overrightarrow{b} \quad \overrightarrow{a} \right ]+0\\ &=\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ]+\left ( -\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ] \right )+\left ( -\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ] \right )\\ &=-\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ] \end{aligned}$

Scalar triple product exercise multiple choice question 15

Answer:
$=3\left [ \overrightarrow{a} \quad \overrightarrow{b}\quad \overrightarrow{c} \right ]$
Hint:
Just simplify the expression
Given:
$\begin{aligned} &=(\overrightarrow{a}+2\overrightarrow{b}-\overrightarrow{c}).\left \{ (\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{c}) \right \}\\ &=(\overrightarrow{a}+2\overrightarrow{b}-\overrightarrow{c}).\left \{ \overrightarrow{a}\times \overrightarrow{a}-\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{a}\times \overrightarrow{c}-\overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c} \right \}\\ &=(\overrightarrow{a}+2\overrightarrow{b}-\overrightarrow{c}).(-\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{c})\\ & =\left [ \overrightarrow{a} \quad \overrightarrow{b}\quad \overrightarrow{c} \right ]+2\left [ \overrightarrow{a} \quad \overrightarrow{b}\quad \overrightarrow{c} \right ]\\ &=3\left [ \overrightarrow{a} \quad \overrightarrow{b}\quad \overrightarrow{c} \right ] \end{aligned}$

Scalar triple product exercise multiple choice question 16

Answer:
$\begin{aligned} &\lambda =-2,\lambda =1\pm \sqrt{3} \end{aligned}$
Hint:
If vectors are coplanar, their scalar triple product is zero.
$\begin{aligned} &\therefore \text { For }\begin{vmatrix} \lambda &1 &2 \\ 1 &\lambda &-1 \\ 2 &-1 &\lambda \end{vmatrix}\\ &=\lambda (\lambda ^2-1)-1(\lambda +2)+2(-1-2\lambda )\\ &=\lambda ^3-\lambda -\lambda -2-2-4\lambda \\ &=\lambda ^3-6\lambda -4 \end{aligned}$
Now
$\begin{aligned} &\lambda ^3-6\lambda -4=0\\ &(\lambda +2)(\lambda ^2-2\lambda -2)=0\\ &\lambda =-2,\lambda =1\pm \sqrt{3} \end{aligned}$

Scalar triple product exercise multiple choice question 17

Answer:
-2
Hint:
Vectors are coplanar, then their scalar triple product is zero
$\begin{aligned} &\therefore \left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ]=\begin{vmatrix} 3 &-1 &2 \\ 2 &1 &3 \\ 1 &\lambda &-1 \end{vmatrix}\\ &=3(-1-3\lambda )+1(-2-3)+2(2\lambda -1)\\ &=-3-9\lambda -2-3+4\lambda -2\\ &=-5\lambda -10 \end{aligned}$
As, vectors are coplanar
$\begin{aligned} &\left [ \overrightarrow{a} \quad \overrightarrow{b} \quad \overrightarrow{c} \right ]=0\\ &-5\lambda -10=0\\ &\lambda =\frac{10}{-5}\\ &\lambda =-2 \end{aligned}$

The concepts in chapter 25, Scalar Triple Product, are challenging compared to the other simple topics. This chapter consists of only one exercise, ex 25.1. There are 17 Multiple Choice Questions asked in this chapter. The concepts that this portion covers include Scalar Triple product formulas, Coplanar, Proof of Scalar Triple Product, and the properties of the Scalar Triple Product. The RD Sharma Class 12 Chapter 25 MCQ material will help the students to clarify their doubts in this part.

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RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

1. How many exercises are present in the class 12 mathematics, chapter 25?

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There are 17 MCQs in the 25th chapter of class 12 mathematics. The RD Sharma book contains the answers to every question in this part.