RD Sharma Class 12 Exercise 25.1 Scalar triple product Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 25.1 Scalar triple product Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 06:36 PM IST

If you are a student who is in class and is looking for a good NCERT solution, your search ends here. The RD Sharma class 12th exercise 25.1 is the best NCERT solutions that students can possibly find today. The book has some high-quality answers which are easy to grasp and understand by all students. The RD Sharma class 12 chapter 25 exercise 25.1 is the perfect book to keep to improve your chances of topping in boards.

RD Sharma class 12 solutions Scalar Triple Product 25.1 should be availed of by all students who have maths in their board exams. Chapter 25 of the NCERT is titled Scalar Triple Product and is a rather short one. RD Sharma solutions The concepts covered are basic ideas on Scalar Triple Product Formula, Volume of parallelopiped, coplanar vectors, Proof of Scalar Triple Product, and Scalar Triple Product Properties. The first exercise of this chapter, ex 25.1 consists of 27 questions along with its subparts. Students can find the solutions for this long list of questions at the RD Sharma Class 12th Exercise 25.1 reference material.

## Scalar Triple Product Excercise: 25.1

Scalar Triple Product Exercise 25.1 Question 1(i)

Hint :- Use formula of scalar triple product
Given:- We know that dot product is commutative.
\begin{aligned} &\therefore[\hat{l} \hat{\jmath} \hat{k}]+[\hat{l} \hat{\jmath} \hat{k}]+[\hat{l} \hat{j} \hat{k}] \\\\ &=3[\hat{l} \hat{\jmath} \hat{k}] \\ \end{aligned}
\begin{aligned} &=3(\hat{\imath} \times \hat{\jmath}) \cdot \hat{k} \\ \end{aligned}(using scalar triple product)
\begin{aligned} &=3(\hat{k}) \cdot \hat{k} \\\\ \end{aligned}
\begin{aligned} &=3(1) \\\ &=3 \end{aligned}

Scalar Triple Product Exercise 25.1 Question 1(ii)

Hint :- Use scalar triple product
Given:- \begin{aligned} &{[2 \hat{\imath} \hat{\jmath} \hat{k}]+[\hat{\imath} \hat{k} \hat{\jmath}]+[\hat{k} \hat{\jmath} 2 \hat{i}]} \\ \end{aligned}
\begin{aligned} &=(2 \hat{\imath} \times \hat{\jmath}) \cdot \hat{k}+(\hat{\imath} \times \hat{k}) \cdot \hat{\jmath}+(\hat{k} \times \hat{\jmath}) \cdot 2 \hat{i} \\ &=2(\hat{k}) \cdot \hat{k}+(-\hat{\jmath}) \cdot \hat{\jmath}+(-\hat{\imath}) \cdot 2 \hat{\imath} \\ &=2(1)+(-1)+(-2) \\ &=2-1-2 \\ &=-1 \end{aligned}

Scalar Triple Product Exercise 25.1 Question 2(i)

Answer: $4$

Hint :- Use cross product & dot product
Given: $\vec{a}=2\hat{i}-3\hat{j}$
$\vec{b}=\hat{i}+\hat{j}-\hat{k}$
$\vec{b}=3\hat{i}-\hat{k}$
We Know $\left [ \vec{a}\vec{b}\vec{c} \right ]=\left ( \vec{a}\times \vec{b} \right ).\vec{c}$
\begin{aligned} &\therefore \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -3 & 0 \\ 1 & 1 & -1 \end{array}\right| \\\\ &=\hat{i}(3-0)-\hat{\jmath}(-2-0)+\hat{k}(2+3) \\\\ &=3 \hat{\imath}+2 \hat{\jmath}+5 \hat{k} \\ \end{aligned}
\begin{aligned} &\text { Now, }(\vec{a} \times \vec{b}) \cdot \vec{c}=(3 \hat{i}+2 \hat{\jmath}+5 \hat{k}) \cdot(3 \hat{i}+0 \hat{j}-\hat{k}) \\\\ &=9+0-5 \\\\ &=4 \end{aligned}

Scalar Triple Product Exercise 25.1 Question 2(ii)

Answer :- $\left [ \vec{a}\:\: \vec{b}\:\: \vec{c} \right ]=12$
Hint :- Use cross product & dot product
Given:$\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$
\begin{aligned} &\vec{b}=2 \hat{\imath}+\hat{\jmath}-\hat{k} \\ &\vec{c}=\hat{\jmath}+\hat{k} \\ \end{aligned}
\begin{aligned} &\text { For, }[\vec{a} \vec{b} \vec{c}]=(\vec{a} \times \vec{b}) \cdot \vec{c} \\ \end{aligned}
\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{array}\right| \\ \end{aligned}
\begin{aligned} &=\hat{l}(2-3)-\hat{\jmath}(-1-6)+\hat{k}(1+4) \\ &=-\hat{\imath}+7 \hat{j}+5 \hat{k} \\ \end{aligned}
\begin{aligned} &(\vec{a} \times \vec{b}) \cdot \vec{c}=(-\hat{\imath}+7 \hat{\jmath}+5 \hat{k}) \cdot(\hat{\jmath}+\hat{k}) \\ &=7+5 \\ &=12 \end{aligned}
$\therefore \left [ \vec{a}\:\: \vec{b}\:\: \vec{c} \right ]=12$

Scalar Triple Product Exercise 25.1 Question 2(iii)

Hint :- Use scalar triple product
Given:- \begin{aligned} &\vec{a}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k} \\ \end{aligned}
\begin{aligned} &\vec{b}=\hat{\imath}-2 \hat{\jmath}+\hat{k} \\ &\vec{c}=-3 \hat{\imath}+\hat{\jmath}+2 \hat{k} \end{aligned}
We Know ,$[\vec{a} \vec{b} \vec{c}]=(\vec{a} \times \vec{b}) \cdot \vec{c}$
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 1 \\ 1 & -2 & 1 \end{array}\right|$
\begin{aligned} &=\hat{\imath}(3+2)-\hat{\jmath}(2-1)+\hat{k}(-4-3) \\\\ &=5 \hat{\imath}-\hat{\jmath}-7 \hat{k} \\ \end{aligned}
Now \begin{aligned} &{[\vec{a} \times \vec{b}] \cdot \vec{c}=[5 \hat{\imath}-\hat{\jmath}-7 \hat{k}) \cdot[-3 \hat{\imath}+\hat{\jmath}+2 \hat{k}]} \\\\ \end{aligned}
\begin{aligned} &\quad=-15-1-14 \\\ \end{aligned}
\begin{aligned} &=-30 \end{aligned}

Scalar Triple Product Exercise 25.1 Question 3(i)

Answer :- 37 cubic units
Hint :- Use formula that volume of parallelepiped is $\left [ \vec{a}\: \vec{b}\: \vec{c} \right ]$
Given:\begin{aligned} &-\vec{a}=2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k} \\ \end{aligned}
\begin{aligned} &\vec{b}=\hat{\imath}+2 \hat{\jmath}-\hat{k} \\ &\vec{c}=3 \hat{\imath}-\hat{\jmath}+2 \hat{k}\ \end{aligned}
\begin{aligned} &{\left[\begin{array}{lll} \vec{a}& \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{ccc} 2 & 3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{array}\right|} \\ \end{aligned}
\begin{aligned} &\quad=2(4-1)-3(2+3)+4(-1-6) \\ \end{aligned}
\begin{aligned} &\quad=2(3)-3(5)+4(-7) \\ \end{aligned}
\begin{aligned} &=6-15-28 \\\ &=-37 \end{aligned}
Volume of parallelepiped,=$\left [ \vec{a}\: \vec{b}\: \vec{c} \right ]$
=$\mid -37\mid$
$=37 \: Cubic \: Units$

Scalar Triple Product Exercise 25.1 Question 3(ii)

Answer :- 35 cubic units
Hint :- Use formula of parallelepiped $=\left [ \vec{a}\: \: \: \vec{b}\: \: \: \vec{c} \right ]$
Given:-\begin{aligned} &-\vec{a}=2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k} \\ \end{aligned}
\begin{aligned} &\vec{b}=\hat{i}+2 \hat{\jmath}-\hat{k} \\ &\vec{c}=3 \hat{\imath}-\hat{\jmath}-2 \hat{k} \\ \end{aligned}
Volume of parallelepiped,$=\left [ \vec{a}\: \: \: \vec{b}\: \: \: \vec{c} \right ]$
\begin{aligned} &=\left|\begin{array}{ccc} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & -2 \end{array}\right| \\ \end{aligned}
\begin{aligned} &=2(-4-1)+3(-2+3)+4(-1-6) \\ &=2(-5)+3(1)+4(-7) \\ &=-10+3-28 \\ &=-35 \\ &=35 \text { cubic units } \end{aligned}

Scalar Triple Product Exercise 25.1 Question 3(iii)

Answer :- 286 cubic units
Hint :- Use formula of volume of parallelepiped
Given:- \begin{aligned} &\vec{a}=11 \hat{\imath}=11 \hat{\imath}+0 \hat{\jmath}+0 \hat{k} \\ \end{aligned}
\begin{aligned} &\vec{b}=2 \hat{\jmath}=0 \hat{\imath}+2 \hat{\jmath}+0 \hat{k} \\ &\vec{c}=13 \hat{k}=0 \hat{\imath}+0 \hat{\jmath}+13 \hat{k} \end{aligned}
Volume of parallelepiped, $=\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]$
\begin{aligned} &=\left|\begin{array}{ccc} 11 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 13 \end{array}\right| \\ &=11(26-0)-0(0-0)+0(0-0) \\ &=286 \text { cubic units } \end{aligned}

Scalar Triple Product Exercise 25.1 Question 3(iv)

Answer :- 4 cubic units
Hint :- Use formula of volume of parallelepiped
Given:\begin{aligned} &\vec{a}=\hat{\imath}+\hat{\jmath}+\hat{k} \\ \end{aligned}
\begin{aligned} &\vec{b}=\hat{i}-\hat{\jmath}+\hat{k} \\ &\vec{c}=\hat{\imath}+2 \hat{\jmath}-\hat{k} \\ \end{aligned}

Volume of parallelepiped,$=\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]$
\begin{aligned} &=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{array}\right| \\ &=1(1-2)-1(-1-1)+1(2+1) \\ &=1(-1)-1(-2)+1(3) \\ &=-1+2+3 \\ &=4 \end{aligned}

Scalar Triple Product Exercise 25.1 Question 4(i)

Answer :- $\left [ \vec{a}\: \: \vec{b}\: \: \vec{c}\right ]=0$
Hint :- vectors are coplanar if there triple scalar product is zero.
Given:-\begin{aligned} &\vec{a}=\hat{\imath}+2 \hat{\jmath}-\hat{k} \\ \end{aligned}
\begin{aligned} &\vec{b}=3 \hat{i}+2 \hat{\jmath}+7 \hat{k} \\ &\vec{c}=5 \hat{\imath}+6 \hat{\jmath}+5 \hat{k} \end{aligned}

For any three vectors, if their scalar triple product is zero. i.e. $\left [ \vec{a}\: \: \vec{b}\: \: \vec{c}\right ]=0$ than they are coplanar.

\begin{aligned} &\therefore\left[\begin{array}{lll} \vec{a} & \vec{b} & c \end{array}\right]=\left|\begin{array}{ccc} 1 & 2 & -1 \\ 3 & 2 & 7 \\ 5 & 6 & 5 \end{array}\right| \\ \end{aligned}
\begin{aligned} &=1(10-42)-2(15-35)-1(18-10) \\ \end{aligned}
\begin{aligned} &= & -32+40-8 \\ \end{aligned}
\begin{aligned} &= & 0 \end{aligned}
Thus, vectors are coplanar.

Scalar Triple Product Exercise 25.1 Question 4(ii)

Answer :- $\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$
Hint :- vectors are coplanar if scalar triple product is zero.
Given:- \begin{aligned} &\vec{a}=\hat{\imath}-2 \hat{\jmath}+3 \hat{k} \\ \end{aligned}
\begin{aligned} &\vec{b}=-2 \hat{\imath}+3 \hat{\jmath}-4 \hat{k} \\ &\vec{c}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k} \end{aligned}
For vectors,$\hat{a},\hat{b}$ &$\hat{c}$ if $\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$. vectors are coplanar.
\begin{aligned} &\therefore\left[\begin{array}{lll} \vec{a} & \vec{b} & c \end{array}\right]=\left|\begin{array}{ccc} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{array}\right| \\ \end{aligned}
\begin{aligned} &=1(15-12)+2(-10+4)+3(6-3) \\ \end{aligned}
\begin{aligned} &= & 3-12+9 \\ \end{aligned}
\begin{aligned} &= & 0 \end{aligned}
Thus, vectors are coplanar.

Scalar Triple Product Exercise 25.1 Question 4(iii)

Answer :- $\left [ \vec{a} \: \: \vec{b} \: \: \vec{c}\right ]=0$
Hint :- vectors are coplanar if scalar triple product is zero.
Given:-\begin{aligned} &\vec{a}=\hat{\imath}-2 \hat{\jmath}+3 \hat{k} \\ \end{aligned}
\begin{aligned} &\vec{b}=-2 \hat{\imath}+3 \hat{\jmath}-4 \hat{k} \\ &\vec{c}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k} \\ \end{aligned}
For vectors, $\hat{a}$$,\hat{b}$ &$\hat{c}$ if $\left [ \vec{a} \: \: \vec{b} \: \: \vec{c}\right ]=0$ vectors are coplanar.
\begin{aligned} &\therefore\left[\begin{array}{ll} \vec{a}\ \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{ccc} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{array}\right| \end{aligned}
$=1\left ( 15-22 \right )+2\left ( -10+4 \right )+3\left ( 6-3 \right )$
$=3-12+9$
$=0$
Thus, vectors are coplanar.

Scalar Triple Product Exercise 25.1 Question 5(i)

Answer :- $\lambda =1$
Hint :- As it is given that vectors are coplanar. So use $\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$ to find $\lambda$.
Given:\begin{aligned} &\vec{a}=\hat{\imath}+\hat{k}-\hat{\jmath} \\ \end{aligned}
\begin{aligned} &\vec{b}=2 \hat{i}+\hat{\jmath}-\hat{k} \\ &\vec{c}=\lambda \hat{\imath}-\hat{\jmath}+\lambda \hat{k} \end{aligned}
As it is mentioned that vectors are coplanar it means that $\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$
\begin{aligned} &\therefore\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ \lambda & -1 & \lambda \end{array}\right| \\ \end{aligned}
\begin{aligned} &=1(\lambda-1)+1(2 \lambda+\lambda)+1(-2-\lambda) \\ &=\lambda-1+3 \lambda-2-\lambda \\ &=3 \lambda-3 \\ \end{aligned}
\begin{aligned} &\text { As, }[\vec{a} \vec{b} \vec{c}]=0 \\ &\therefore 3 \lambda-3=0 \\ &3 \lambda=3 \\ &\lambda=1 \end{aligned}

Scalar Triple Product Exercise 25.1 Question 5(ii)

Answer :- $\lambda =\frac{-25}{8}$
Hint :- As it is given that vectors are coplanar. So use $\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$ to find $\lambda$
Given:- \begin{aligned} &\vec{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k} \\ \end{aligned}
\begin{aligned} &\vec{b}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k} \\ &\vec{c}=\lambda \hat{\imath}+\lambda \hat{j}+5 \hat{k} \end{aligned}

As it is given vectors are coplanar

$\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$
$=\left|\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -3 \\ \lambda & \lambda & 5 \end{array}\right|$
\begin{aligned} =2(10+3 \lambda)+1(5+3 \lambda)+1(\lambda-2 \lambda) \end{aligned}
\begin{aligned} &=20+6 \lambda+5+3 \lambda-\lambda \\ &=8 \lambda+25 \\ &\text { As, }[\vec{a} \vec{b} \vec{c}]=0 \\ &\therefore 8 \lambda+25=0 \\ &8 \lambda=-25 \\ &\lambda=\frac{-25}{8} \end{aligned}

Scalar Triple Product Exercise 25.1 Question 5(iii)

Answer:$\lambda =6$
Hint :- As it is given that vectors are coplanar. So use $\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$ to find $\lambda$.
Given:\begin{aligned} &\vec{a}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k} \\ \end{aligned}
\begin{aligned} &\vec{b}=3 \hat{\imath}+\lambda \hat{\jmath}+\hat{k} \\ &\vec{c}=\hat{\imath}+2 \hat{\jmath}+2 \hat{k} \end{aligned}
As it is mentioned that vectors are coplanar, it means that $\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$
\begin{aligned} &=\left|\begin{array}{ccc} 1 & 2 & -3 \\ 3 & \lambda & 1 \\ 1 & 2 & 2 \end{array}\right| \\ &=1(2 \lambda-2)-2(6-1)-3(6-\lambda) \\ &=2 \lambda-2-10-18+3 \lambda \\ &=5 \lambda-30 \end{aligned}
Now,$\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$
$\therefore 5\lambda -30=0$
$5\lambda =30$
$\lambda =6$

Scalar Triple Product Exercise 25.1 Question 5(iv)

Answer :- $\lambda =\frac{-1}{3}$
Hint :- As it is given that vectors are coplanar. So use $\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$ to find $\lambda$
Given: \begin{aligned} &\vec{a}=\hat{\imath}+3 \hat{\jmath}=\hat{i}+3 \hat{\jmath}+0 \hat{k} \\ \end{aligned}
\begin{aligned} &\vec{b}=5 \hat{k}=0 \hat{\imath}+0 \hat{\jmath}+5 \hat{k} \\ &\vec{c}=\lambda \hat{\imath}-\hat{\jmath}=\lambda \hat{\imath}-\hat{\jmath}+0 \hat{k} \end{aligned}
If vectors are given coplanar then,
$\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$
\begin{aligned} &{\left[\begin{array}{ll} \vec{a}\ \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{ccc} 1 & 3 & 0 \\ 0 & 0 & 5 \\ \lambda & -1 & 0 \end{array}\right|} \\ \end{aligned}
\begin{aligned} &=1(0+5)-3(0-5 \lambda)+0(0-0) \\ &=5+15 \lambda \\ \end{aligned}
As $\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]=0$
\begin{aligned} &5+15 \lambda=0 \\ &\therefore \lambda=\frac{-5}{15} \\ &\lambda=\frac{-1}{3} \end{aligned}

Scalar Triple Product Exercise 25.1 Question 6

Answer :- $\left [ \vec{AB}\: \: \vec{AC}\: \: \vec{AD} \right ]\neq 0$
Hint :- if any triads of vectors are not coplanar if & only if their scalar triple product $\neq 0$
Given:- four points A,B,C & D
So, let \begin{aligned} &\vec{A}=6 \hat{\imath}-7 \hat{\jmath} \\ \end{aligned}
\begin{aligned} &\vec{B}=16 \hat{\imath}-19 \hat{j}-4 \hat{k} \\ &\vec{C}=3 \hat{\imath}-6 \hat{k} \\ &\vec{D}=2 \hat{\imath}-5 \hat{j}+10 \hat{k} \end{aligned}
We have to show that any three vectors forming from four points have their scalar triple product$\neq 0$
$\therefore \vec{AB}=$ Position of B-Position of A
\begin{aligned} &=(16 \hat{\imath}-19 \hat{\jmath}-4 \hat{k})-(6 \hat{\imath}-7 \hat{\jmath}+0 \hat{k}) \\ &=(16-6) \hat{\imath}+(-19+7) \hat{\jmath}+(-4-0) \hat{k} \\ &=10 \hat{\imath}-12 \hat{\jmath}-4 \hat{k} \end{aligned}
Similarly, $\therefore \vec{AC}=$ Position of C -Position of A
\begin{aligned} &=(3 \hat{\imath}+0 \hat{\jmath}-6 \hat{k})-(6 \hat{\imath}-7 \hat{\jmath}+0 \hat{k}) \\ &=(3-6) \hat{\imath}+(0+7) \hat{\jmath}+(-6-0) \hat{k} \\ &=-3 \hat{\imath}+7 \hat{\jmath}-6 \hat{k} \end{aligned}
Now, $\therefore \vec{AD}=$ Position of D – Position of A
\begin{aligned} &=(2 \hat{\imath}-5 \hat{\jmath}+10 \hat{k})-(6 \hat{\imath}-7 \hat{\jmath}+0 \hat{k}) \\ &=(2-6) \hat{\imath}+(-5+7) \hat{\jmath}+(10-0) \hat{k} \\ &=-4 \hat{\imath}+2 \hat{\jmath}+10 \hat{k} \end{aligned}
Now, we have to prove.$\left [ \vec{AB}.\vec{AC}.\vec{AD} \right ]\neq 0$
\begin{aligned} &\therefore[\overrightarrow{A B} \cdot \overrightarrow{A C} \cdot \overrightarrow{A D}]=\left|\begin{array}{ccc} 10 & -12 & -4 \\ -3 & 7 & -6 \\ -4 & 2 & 10 \end{array}\right| \\ &=10(70+12)+12(-30-24)-4(-6+28) \\ &=820-648-88 \\ &=84 \neq 0 \end{aligned}

Thus, the vectors so formed are not coplanar.

Scalar Triple Product Exercise 25.1 Question 7

Answer :- $\left [ \vec{AB}\: \: \vec{AC}\: \: \vec{AD} \right ]=0$
Hint :- To prove vectors are coplanar scalar product of three vectors should be zero.
Given:- points $A=\hat{i}+4\hat{j}-3\hat{k}$\
$B=3\hat{i}+2\hat{j}-5\hat{k}$
$C=3\hat{i}+8\hat{j}-5\hat{k}$
$D=3\hat{i}+2\hat{j}+\hat{k}$
\begin{aligned} &\therefore \overrightarrow{A B}=(3 \hat{\imath}+2 \hat{\jmath}-5 \hat{k})-(-\hat{\imath}+4 \hat{\jmath}-3 \hat{k}) \\ &=(3+1) \hat{\imath}+(2-4) \hat{\jmath}+(-5+3) \hat{k} \\ &=4 \hat{\imath}-2 \hat{\jmath}-2 \hat{k} \end{aligned}
Similarly, \begin{aligned} &\overrightarrow{A C}=(-3 \hat{\imath}+8 \hat{\jmath}-5 \hat{k})-(-\hat{\imath}+4 \hat{\jmath}-3 \hat{k}) \\ \end{aligned}
\begin{aligned} &=(-3+1) \hat{\imath}+(8-4) \hat{\jmath}+(-5+3) \hat{k} \\ &=-2 \hat{\imath}+4 \hat{\jmath}-2 \hat{k} \\ &\overrightarrow{A D}=(-3 \hat{\imath}+2 \hat{\jmath}+\hat{k})-(-\hat{\imath}+4 \hat{\jmath}-3 \hat{k}) \\ &=(-3+1) \hat{\imath}+(2-4) \hat{\jmath}+(1+3) \hat{k} \\ &=-2 \hat{\imath}-2 \hat{\jmath}+4 \hat{k} \end{aligned}
Now, if $\left [ \vec{AB}\: \: \vec{AC}\: \: \vec{AD} \right ]=0$ Vectors are coplanar
\begin{aligned} &\therefore[\overrightarrow{A B} \overrightarrow{A C} \overrightarrow{A D}]=\left|\begin{array}{ccc} 4 & -2 & -2 \\ -2 & 4 & -2 \\ -2 & -2 & 4 \end{array}\right| \\ \end{aligned}
\begin{aligned} &=4(16-4)+2(-8-4)-2(4+8) \\ &=4(12)+2(-12)-2(12) \\ &=48-24-24 \\ &=0 \end{aligned}
Thus given vectors are coplanar.

Scalar Triple Product Exercise 25.1 Question 8

Answer :- $\left [ \overrightarrow{AB}\: \overrightarrow{AC}\: \overrightarrow{AD} \right ]\neq 0$
Hint :- To show vectors are coplanar, their scalar triple product must be zero.
Given:- points whose position vectors are,
\begin{aligned} &\vec{A}=6 \hat{\imath}-7 \hat{\jmath} \\ &\vec{B}=16 \hat{\imath}-19 \hat{\jmath}-4 \hat{k} \\ &\vec{C}=3 \hat{\imath}-6 \hat{k} \\ &\vec{D}=2 \hat{\imath}-5 \hat{\jmath}+10 \hat{k} \end{aligned}
\begin{aligned} &\overrightarrow{A B}=(16 \hat{\imath}-19 \hat{\jmath}-4 \hat{k})-(6 \hat{\imath}-7 \hat{\jmath}) \\ &=(16-6) \hat{\imath}+(-19+7) \hat{\jmath}+(-4-0) \hat{k} \\ &=10 \hat{\imath}-12 \hat{\jmath}-4 \hat{k} \\ \end{aligned}
\begin{aligned} &\overrightarrow{A C}=(3 \hat{\imath}-6 \hat{k})-(6 \hat{\imath}-7 \hat{\jmath}) \\ &=(3-6) \hat{\imath}+(0+7) \hat{\jmath}+(-6-0) \hat{k} \\ &=-3 \hat{\imath}+7 \hat{\jmath}-6 \hat{k} \\ \end{aligned}
\begin{aligned} &\overrightarrow{A D}=(2 \hat{\imath}-5 \hat{j}+10 \hat{k})-(6 \hat{\imath}-7 \hat{\jmath}) \\ &=(2-6) \hat{\imath}+(-5+7) \hat{\jmath}+(10-0) \hat{k} \\ &=-4 \hat{\imath}+2 \hat{\jmath}+10 \hat{k} \end{aligned}
Now, we have to prove. $\left [ \overrightarrow{AB}\: \overrightarrow{AC}\: \overrightarrow{AD} \right ]\neq 0$
\begin{aligned} &\therefore[\overrightarrow{A B} \overrightarrow{A C} \overrightarrow{A D}]=\left|\begin{array}{ccc} 10 & -12 & -4 \\ -3 & 7 & -6 \\ -4 & 2 & 10 \end{array}\right| \\ \end{aligned}
\begin{aligned} &=10(70+12)+12(-30-24)-4(-6+28) \\ &=820-648+(-88) \\ &=84 \neq 0 \end{aligned}
Thus, the vectors are not coplanar.

Scalar Triple Product Exercise 25.1 Question 9

Answer :- $\lambda =1$
Hint :- When vectors are coplanar their scalar triple product is zero.
Given: four points,
\begin{aligned} &\mathrm{A}=-\hat{\jmath}-\hat{k} \\ &\mathrm{~B}=4 \hat{\imath}+5 \hat{\jmath}+\lambda \hat{k} \\ &\mathrm{C}=3 \hat{\imath}+9 \hat{\jmath}+4 \hat{k} \\ &\mathrm{D}=-4 \hat{\imath}+4 \hat{\jmath}+4 \hat{k} \\ \end{aligned}
\begin{aligned} &\therefore \overrightarrow{A B}=(4 \hat{\imath}+5 \hat{\jmath}+\lambda \hat{k})-(-\hat{\jmath}-\hat{k}) \\ &=(4-0) \hat{\imath}+(5+1) \hat{\jmath}+(\lambda+1) \hat{k} \\ &=4 \hat{\imath}+6 \hat{\jmath}+(\lambda+1) \hat{k} \end{aligned}
\begin{aligned} &\overrightarrow{A C}=(3 \hat{\imath}+9 \hat{\jmath}+4 \hat{k})-(-\hat{\jmath}-\hat{k}) \\ &=(3-0) \hat{\imath}+(9+1) \hat{\jmath}+(4+1) \hat{k} \\ &=3 \hat{\imath}+10 \hat{\jmath}+5 \hat{k} \\ \end{aligned}
\begin{aligned} &\overrightarrow{A D}=(-4 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})-(-\hat{\jmath}-\hat{k}) \\ &=(-4-0) \hat{\imath}+(4+1) \hat{\jmath}+(4+1) \hat{k} \\ &=-4 \hat{\imath}+5 \hat{\jmath}+5 \hat{k} \end{aligned}

Now, As vectors are given coplanar then,$\left [ \overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD} \right ]=0$

\begin{aligned} &\therefore[\overrightarrow{A B} \overrightarrow{A C} \overrightarrow{A D}]=\left|\begin{array}{ccc} 4 & 6 & (\lambda+1) \\ 3 & 10 & 5 \\ -4 & 5 & 5 \end{array}\right| \\ \end{aligned}

\begin{aligned} &=4(50-25)-6(15+20)+(\lambda+1)(15+40) \\ &=100-210+55 \lambda+55 \\ &=55 \lambda+(-55) \\ &=55 \lambda-55 \\ \end{aligned}

As vectors are coplanar

\begin{aligned} &\therefore 55 \lambda-55=0 \\ &\therefore \lambda=1 \end{aligned}

Scalar Triple Product Exercise 25.1 Question 10

Answer:$\left ( \vec{a}-\vec{b} \right ).\left \{ \left ( \vec{b}-\vec{c} \right )\times \left ( \vec{c}-\vec{a} \right ) \right \}=0$
Hint :- Use cross-product to prove
Given:$\left ( \vec{a}-\vec{b} \right ).\left \{ \left ( \vec{b}-\vec{c} \right )\times \left ( \vec{c}-\vec{a} \right ) \right \}=0$
\begin{aligned} \text { L.H.S } &=\left(\vec{a}_{-} \vec{b}\right) \cdot\{(\vec{b}-\vec{c}) \times(\vec{c}-\vec{a})\} \\ &=(\vec{a}-\vec{b}) \cdot\{\vec{b} \times(\vec{c}-\vec{a})-\vec{c} \times(\vec{c}-\vec{a})\} \\ \end{aligned}
\begin{aligned} &=(\vec{a}-\vec{b}) \cdot((\vec{b} \times \vec{c})-(\vec{b} \times \vec{a})-(\vec{c} \times \vec{c})+(\vec{c} \times \vec{a})) \: \: \: \quad(\because \text { Distributive law }) \\ \end{aligned}
\begin{aligned} &=(\vec{a}-\vec{b}) \cdot((\vec{b} \times \vec{c})-(\vec{b} \times \vec{a})-0+(\vec{c} \times \vec{a})) \quad(\because \vec{c} \times \vec{c}=0) \\\\ &=(\vec{a}-\vec{b}) \cdot((\vec{b} \times \vec{c})+(\vec{a} \times \vec{b})+(\vec{c} \times \vec{a})) \quad(\vec{a}) \\\\ &=\vec{a} \cdot(\vec{b} \times \vec{c})+\vec{a} \cdot(\vec{a} \times \vec{b})+\vec{a} \cdot(\vec{c} \times \vec{a})-\vec{b} \cdot(\vec{b} \times \vec{c})-\vec{b} \cdot(\vec{a} \times \vec{b})-\vec{b} \cdot(\vec{c} \times \vec{a}) \\\\ &=\left[\begin{array}{ll} \vec{a} \vec{b} & \vec{c}]+0+0-0-0-[\vec{a} \vec{b} \vec{c}] \end{array}\right.\\\\ &=0 \\ &=\mathrm{R} . \mathrm{H} . \mathrm{S} \text { (Hence proved) } \end{aligned}

Scalar Triple Product Exercise 25.1 Question 11

write the ques number in proper format
Hint :- When vectors are perpendicular their dot product is zero.
Given:- $\vec{a},\vec{b}$ & $\vec{c}$ be position vectors of A, B & C
\begin{aligned} &\therefore \overrightarrow{A B}=(\vec{b}-\vec{a}) \\ &\overrightarrow{B C}=(\vec{c}-\vec{b}) \\ &\overrightarrow{C A}=(\vec{a}-\vec{c}) \end{aligned}

If a vector given $\left ( \vec{a}\times \vec{b} \right )+\left ( \vec{b}\times \vec{c} \right )+\left ( \vec{c}\times \vec{a} \right )$ is perpendicular to the plane of triangle ABC then its dot product with the vector $\overrightarrow{AB},\overrightarrow{BC},\overrightarrow{CA}$ must be zero

\begin{aligned} &\overrightarrow{A B} \cdot[(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})]\\ &=[(\vec{b}-\vec{a}) \cdot(\vec{a} \times \vec{b})]+[(\vec{b}-\vec{a}) \cdot(\vec{b} \times \vec{c})]+[(\vec{b}-\vec{a}) \cdot(\vec{c} \times \vec{a})]\\ \end{aligned}

\begin{aligned} &=[(\vec{b} \cdot(\vec{a} \times \vec{b})]-[\vec{a} \cdot(\vec{a} \times \vec{b})]+[\vec{b} \cdot(\vec{b} \times \vec{c})]-[\vec{a} \cdot(\vec{b} \times \vec{c})]+[(\vec{b} \cdot(\vec{c} \times \vec{a})]-[\vec{a} \cdot(\vec{c} \times \vec{a})]\\ \end{aligned}

$\inline =\left[\begin{array}{lll} \vec{b} & \vec{a} & \vec{b} \end{array}\right]-\left[\begin{array}{lll} \vec{a} & \vec{a} & \vec{b} \end{array}\right]+\left[\begin{array}{lll} \vec{b} & \vec{b} & \vec{c} \end{array}\right]-\left[\begin{array}{lll} \vec{a} & \vec{b} & c \end{array}\right]+\left[\begin{array}{lll} \vec{b} & \vec{c} & a \end{array}\right]-\left[\begin{array}{lll} \vec{a} & \vec{c} & \vec{a} \end{array}\right]$

\inline \begin{aligned} &=0-0+0-[\vec{a} \vec{b} \vec{c}]+[\vec{a} \vec{b} \vec{c}]-0 \\ &=0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad(\because[\vec{b} \vec{c} \vec{a}]=[\vec{a} \vec{b} \vec{c}]) \end{aligned}

\inline \begin{aligned} &\overrightarrow{B C} \cdot[(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})] \\\\ &=[\vec{c}-\vec{b}][(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})] \end{aligned}

$\inline =\left[\begin{array}{lll} \vec{c} & \vec{a} & \vec{b} \end{array}\right]-\left[\begin{array}{lll} \vec{b} & \vec{a} & \vec{b} \end{array}\right]+\left[\begin{array}{lll} \vec{c} & \vec{b} & \vec{c} \end{array}\right]-\left[\begin{array}{lll} \vec{b} & \vec{b} & \vec{c} \end{array}\right]+\left[\begin{array}{lll} \vec{c} & \vec{c} & \vec{a} \end{array}\right]-\left[\begin{array}{lll} \vec{b} & \vec{c} & \vec{a} \end{array}\right]$

$\inline =\left [ \vec{c}\: \: \vec{a}\: \: \vec{b} \right ]+0+0-0-0-\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]$

$\inline =0$ $\inline =\left ( \because \left [ \vec{c}\: \: \vec{a}\: \: \vec{b} \right ]=\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ] \right )$

Similarly,

\inline \begin{aligned} &\overrightarrow{C A} \cdot[(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})] \\\\ &=[\vec{a}-\vec{c}][(\vec{a} \times \vec{b})+(\vec{b} \times \vec{c})+(\vec{c} \times \vec{a})] \end{aligned}

$\inline =\left[\begin{array}{lll} \vec{a} & \vec{a} & \vec{b} \end{array}\right]-\left[\begin{array}{lll} \vec{c} & \vec{a} & \vec{b} \end{array}\right]+\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]-\left[\begin{array}{lll} \vec{c} & \vec{b} & \vec{c} \end{array}\right]+\left[\begin{array}{lll} \vec{a} \vec{c} & \vec{a} \end{array}\right]-\left[\begin{array}{lll} \vec{c} & \vec{a} \end{array}\right]$

$\inline =0-\left [ \vec{c}\: \: \vec{a}\: \: \vec{b} \right ]\left [ \vec{c}\: \vec{a}\: \vec{b} \ \right ]-0+0-0$

$\inline =0$
As, dot product of all vectors is zero. We can say that $\left ( \vec{a}\times \vec{b} \right )+\left ( \vec{b}\times \vec{c} \right )+\left ( \vec{c}\times \vec{a} \right )$ is perpendicular to the plane of triangle ABC.

Scalar Triple Product Exercise 25.1 Question 12(i)

Answer:$c_{3}=2$
Hint :- If vectors are coplanar their scalar triple product is zero.
Given:$\vec{a}=\hat{i}+\hat{j}+\hat{k}$

$\vec{b}=\hat{i}$

\begin{aligned} &\vec{c}=\mathrm{c}_{1} \hat{\imath}+\mathrm{c}_{2} \hat{\jmath}+\mathrm{c}_{3} \hat{k} \\ \end{aligned}

If c 1 = 1 & c2 = 2 then find c3 which makes $\vec{a}$ , $\vec{b}$ & $\vec{c}$ are coplanar.

\begin{aligned} &\therefore \vec{c}=1 \vec{\imath}+2 \vec{\jmath}+\mathrm{c}_{3} \vec{k} \\ \end{aligned}

If $\vec{a}$,$\vec{b}$,$\vec{c}$ are coplanar than $\left [ \vec{a}\: \vec{b}\: \vec{c} \right ]=0$

\begin{aligned} &\therefore\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & c_{3} \end{array}\right|=0 \\ &1(0-0)-1\left(c_{3}-0\right)+1(2-0)=0 \\ &-c_{3}+2=0 \\ &\therefore-c_{3}+2=0 \\ &c_{3}=2 \end{aligned}

Scalar Triple Product Exercise 25.1 Question 12(ii)

Answer:- no value of $c_{1}$ can make $\vec{a},\vec{b},\vec{c}$ coplanar
Hint :- If three vectors are coplanar than scalar triple product is zero.
Given:-$\vec{a}=\hat{i}+\hat{j}+\hat{k}$
$\vec{b}=\hat{i}$
\begin{aligned} &\vec{c}=c_{1} \hat{\imath}+c_{2} \hat{\jmath}+c_{3} \hat{k} \\ \end{aligned}
Also,\begin{aligned} &c_{2}=-1, c_{3}=1 \\ \end{aligned}
\begin{aligned} &\therefore \vec{c}=c_{1} \vec{l}-\vec{J}+\vec{k} \end{aligned}

For $\vec{a},\vec{b},\vec{c}$to be co-planar $\left [ \vec{a},\vec{b},\vec{c} \right ]=0$

$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & 0 \\ c_{1} & -1 & 1 \end{array}\right|$

$=1\left ( 0-0 \right )-1\left ( 1-0 \right )+1\left ( -1-0 \right )$

$=0-1-1$

$=-2$

But $=-2\neq 0$

And no value of $c_{1}$ will be enough to change the criteria. Thus no value of $c_{1}$can make$\vec{a},\vec{b},\vec{c}$ coplanar.

Scalar Triple Product Exercise 25.1 Question 13

Answer :- $\lambda =5$
Hint :- If the shown are coplanar, their scalar triple product is zero.
Given:- points
\begin{aligned} &\mathrm{A}=3 \hat{\imath}+2 \hat{\jmath}+\hat{k} \\ &\mathrm{~B}=4 \hat{\imath}+\lambda \hat{\jmath}+5 \hat{k} \\ &\mathrm{C}=4 \hat{\imath}+2 \hat{\jmath}-2 \hat{k} \\ &\mathrm{D}=6 \hat{\imath}+5 \hat{\jmath}-\hat{k} \end{aligned}
Vectors formed by these points are coplanar.
So, we will form any three vectors & their scalar triple product will be zero & will get the value of $\lambda$.
\begin{aligned} &\therefore \overrightarrow{A B}=(4 \hat{\imath}+\lambda \hat{\jmath}+5 \hat{k})-(3 \hat{\imath}+2 \hat{\jmath}+\hat{k}) \\ &=(4-3) \hat{\imath}+(\lambda-2) \hat{\jmath}+(5-1) \hat{k} \\ &=\hat{\imath}+(\lambda-2) \hat{\jmath}+4 \hat{k} \\ \end{aligned}
\begin{aligned} &\overrightarrow{A C}=(4 \hat{\imath}+2 \hat{\jmath}-2 \hat{k})-(3 \hat{\imath}+2 \hat{\jmath}+\hat{k}) \\ &=(4-3) \hat{\imath}+(2-2) \hat{\jmath}+(-2-1) \hat{k} \end{aligned}
$=\hat{i}-3\hat{k}$
\begin{aligned} &\overrightarrow{A D}=(6 \hat{\imath}+5 \hat{\jmath}-\hat{k})-(3 \hat{\imath}+2 \hat{\jmath}+\hat{k}) \\ &=(6-3) \hat{\imath}+(5-2) \hat{\jmath}+(-1-1) \hat{k} \\ &=3 \hat{\imath}+3 \hat{\jmath}-2 \hat{k} \end{aligned}
\begin{aligned} &{[\overrightarrow{A B} \overrightarrow{A C} \overrightarrow{A D}]=0} \\ &=\left|\begin{array}{ccc} 1 & (\lambda-2) & 4 \\ 1 & 0 & -3 \\ 3 & 3 & -2 \end{array}\right| \end{aligned}
\begin{aligned} &=1(0+9)-(\lambda-2)(-2+9)+4(3-0) \\ &=9-(\lambda-2)(7)+12 \\ &=9-7 \lambda+14+12 \\ \end{aligned}
\begin{aligned} &=35-7 \lambda \\ &\therefore 35-7 \lambda=0 \\ &\therefore 7\lambda=35 \\ &\lambda=5 \end{aligned}

Scalar Triple Product Exercise 25.1 Question 14

Answer :- $x=6$
Hint :- If given vectors are coplanar, than their scalar triple product is always zero.
Given:- \begin{aligned} &\mathrm{A}=4 \hat{\imath}+3 \hat{\jmath}+3 \hat{k} \\ \end{aligned}
\begin{aligned} &\mathrm{B}=5 \hat{\imath}+x \hat{\jmath}+7 \hat{k} \\ &\mathrm{C}=5 \hat{\imath}+3 \hat{\jmath} \\ &\mathrm{D}=7 \hat{\imath}+6 \hat{j}+\hat{k} \end{aligned}
We have to form any three vectors from above points & make their scalar triple product is zero to find the value of x.
\begin{aligned} &\therefore \overrightarrow{A B}=(5 \hat{\imath}+x \hat{\jmath}+7 \hat{k})-(4 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}) \\ &=(5-4) \hat{\imath}+(\mathrm{x}-3) \hat{\jmath}+(7-3) \hat{k} \\ &=\hat{\imath}+(\mathrm{x}-3) \hat{\jmath}+4 \hat{k} \\ \end{aligned}
\begin{aligned} &\overrightarrow{A C}=(5 \hat{\imath}+3 \hat{\jmath})-(4 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}) \\ &=(5-4) \hat{\imath}+(3-3) \hat{\jmath}+(0-3) \hat{k} \end{aligned}
$=\hat{i}-3\hat{k}$
\begin{aligned} &\overrightarrow{A D}=(7 \hat{\imath}+6 \hat{\jmath}+\hat{k})-(4 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}) \\ &=(7-4) \hat{\imath}+(6-3) \hat{\jmath}+(1-3) \hat{k} \\ &=3 \hat{\imath}+3 \hat{\jmath}-2 \hat{k} \end{aligned}
We know $\left [ \overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD} \right ]=0$
$\therefore\left|\begin{array}{ccc} 1 & (x-3) & 4 \\ 1 & 0 & -3 \\ 3 & 3 & -2 \end{array}\right|$
\begin{aligned} &=1(0+9)-(\mathrm{x}-3)(-2+9)+4(3-0) \\ &=9-7 \mathrm{x}+21+12 \\ &=42-7 \mathrm{x} \\ &\therefore 42-7 \mathrm{x}=0 \\ &\therefore 7 \mathrm{x}=42 \\ &\mathrm{x}=6 \end{aligned}

Scalar Triple Product Exercise 25.1 Question 15

Answer :- 24 cubic units
Hint :- Use formula of volume of parallelopiped $=\left [ \vec{a}\: \: \vec{b}\: \: \vec{c} \right ]$ where, $\vec{a}\: \: \vec{b}$ & $\: \: \vec{c}$ be the adjacent edges.
Given:\begin{aligned} &2 \vec{a}=2(\hat{\imath}-\hat{\jmath}+2 \hat{k})=2 \hat{\imath}-2 \hat{\jmath}+4 \hat{k} \\ \end{aligned}
\begin{aligned} &-\vec{b}=-(3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k})=-3 \hat{\imath}-4 \hat{\jmath}+5 \hat{k} \\ &3 \vec{c}=3(2 \hat{\imath}-\hat{\jmath}+3 \hat{k})=6 \hat{\imath}-3 \hat{\jmath}+9 \hat{k} \end{aligned}
∴ Volume of parallelopiped $=\mid \left [ 2\vec{a}\left ( -\vec{b} \right ) \left ( 3\vec{c} \right )\right ]\mid$
$=\left|\begin{array}{ccc} 2 & -2 & 4 \\ -3 & -4 & 5 \\ 6 & -3 & 9 \end{array}\right|$
\begin{aligned} &=2(-36+15)+2(-27-30)+4(9+24) \\\\ &=2(-21)+2(-57)+4(33) \\\\ &=-42-114+132 \\\\ &=-24 \end{aligned}
∴ Volume of parallelopiped $=\mid -24\mid =24$ cubic units .

The class 12 RD Sharma chapter 25 exercise 25.1 solution is the ideal book to use if students want to perfect their maths skills. Here are some excellent benefits of using the RD Sharma class 12th exercise 25.1:-

• RD Sharma class 12th exercise 25.1 can be of great help to students when they practice at home. They can use the solutions in the book to check their own answers and correct their mistakes. Students might even find common questions in boards if they are thorough with their studies.

• RD Sharma class 12 solutions chapter 25 ex 25.1 is used by teachers to give home to students. Therefore, students can use the book to answer tough homework questions.

• Experts in mathematics have crafted the answers in the book with much care. They have used some unique methods of calculation which will be helpful for all students in enhancing their skills.

• RD Sharma class 12th exercise 25.1 has an updated syllabus which is changed with the latest version of the NCERT maths textbooks. You will be able to download the latest version of the book from the Career360 website.

• There are many expensive options in NCERT solutions which are rarely helpful. The RD Sharma class 12th exercise 25.1 is beneficial for students and comes free of cost.

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

RD Sharma Chapter wise Solutions

### Frequently Asked Question (FAQs)

1. How does the RD Sharma books help the class 12 students who are preparing for JEE mains?

The RD Sharma Class 12th Exercise 25.1 solution books include an abundance of solved sums and practice questions. This makes the students attain a better practice while they are preparing for their JEE mains.

2. What makes the RD Sharma Class 12 Solutions Scalar Triple Product Ex 25.1 Solutions, the most trusted reference guide?

The solutions given in the RD Sharma Class 12th Exercise 25.1 reference guide are crafted by a team of experts. When guidance is provided by an expert group, the students' trust on the answer key increases proportionally. And eventually, many 12th graders utilize it.

3. What is a better way to download the RD Sharma Solution books costless?

These RD Sharma books can be downloaded without any payment from the career 360 website.

4. Write the concepts that are covered in RD Sharma Class 12 chapter 25 in Mathematics?

RD Sharma class 12 chapter 25 in Mathematics contains concepts under the topic Algebra of Vectors. The chapter includes the basic ideas on Scalar Triple Product Formula, Proof of Scalar Triple Product, and Scalar Triple Product Properties.

5. Can I use RD Sharma Solutions for homework?

School teachers often give questions from RD Sharma Solutions to test students. Students can easily find the answers to their homework questions in the RD Sharma solutions.

Get answers from students and experts