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

# 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.

## Scalar Triple Product Excercise: 25 MCQ

Scalar triple product exercise multiple choice question 1

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

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

$\pm 1$
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

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

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

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

$=\frac{\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2}{4}$
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

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:
\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

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

$=\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2$
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

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

$\overrightarrow{a}.\overrightarrow{b}=0,\overrightarrow{b}.\overrightarrow{c}=0,\overrightarrow{c}.\overrightarrow{a}=0$
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

\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

\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

$=3\left [ \overrightarrow{a} \quad \overrightarrow{b}\quad \overrightarrow{c} \right ]$
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

\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

-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.

Not every student can work out the sums efficiently without good practice. The main advantage of the RD Sharma Class 12th Chapter 25 MCQ reference guide is that numerous additional solved Multiple Choice Questions are present. When the students work out all these questions, there is a high chance that their understanding will increase. As the Class 12 RD Sharma Chapter 25 MCQ Solution guide follows the NCERT pattern, the students of the CBSE schools get with it quickly.

The students of class 12 use the RD Sharma books to complete their homework and home assignments. This reference material is also helpful to make them prepare for various tests and exams, including the public exams and the JEE mains exam. Just with the help of RD Sharma Class 12 Solutions Scalar Triple Product Chapter 25 MCQ, they can face any challenging questions effortlessly. RD Sharma Solutions Therefore, they can clarify their doubts without the help of a teacher or a tutor. This is more beneficial for the students who prefer to stay away from extra tuition.

The professionals who have worked in framing the answer key for this portion have expertise in the mathematical domain. The RD Sharma Class 12th Chapter 25 MCQ reference material is the strongly recommended mathematics guide by the previous batch students. There are high chances that the upcoming public exam questions paper will contain questions picked from the RD Sharma books.

Moreover, the career 360 website provides completely free of cost access and download options for its visitors. This attracts many students to buy the RD Sharma Class 12 Solutions Chapter 25 MCQ as they need not spend a lump sum on other reference books.

RD Sharma Chapter wise Solutions

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

Class 12, chapter 25 consists of only one exercise, ex 25.1 in the textbook. To view the solved sums for this exercise and the other sections, refer to the RD Sharma Class 12th Chapter 25 MCQ book.

2. What is the minimum monetary charge that one has to pay to access the RD Sharma books?

3. What is the most recommended solution book to clarify the doubts regarding the MCQs in mathematics, chapter 25?

The RD Sharma Class 12th Chapter 25 MCQ solution material is the most recommended guide to clarify the doubts regarding the chapter 25 multiple-choice questions.

4. Can everyone access the RD Sharma books on the Career 360 website for free?

There are no restrictions on who can access the RD Sharma books on the Career 360 website. Class 12 students, teachers, tutors, parents, and anyone who visits the site can view the solution books for free.

5. How many MCQs are answered in chapter 25 of the RD Sharma solution book?

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.

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