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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.

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

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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

Updated on 25 Jan 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.

RD Sharma Class 12 Solutions Chapter25 Scalar Triple Product - Other Exercise

Scalar Triple Product Excercise: 25.1

Scalar Triple Product Exercise 25.1 Question 1(i)

Answer :- 3
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)

Answer :- -1
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)

Answer :- -30
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
Answer??
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

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

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

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

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

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

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

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

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

Similarly,

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

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

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

$=0$
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 c₃which 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 .

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