Access to comprehensive Understanding ISC Mathematics Class 12 Solutions Chapter 1 Vectors Ex 1.4 encourages independent learning.

## ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.4

Question 1.
If $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$$, $$\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}$$ and $$\vec{c}=3 \hat{i}+4 \hat{j}-\hat{k}$$, then find $$(\vec{a} \times \vec{b}) \cdot \vec{c}$$ and $$\vec{a} \cdot(\vec{b} \times \vec{c})$$. Is $$(\vec{a} \times \vec{b}) \cdot \vec{c}=\vec{a} \cdot(\vec{b} \times \vec{c})$$ ?
Solution:
Given $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$$,
$$\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}$$
and $$\vec{c}=3 \hat{i}+4 \hat{j}-\hat{k}$$

$$(\vec{a} \times \vec{b}) \cdot \vec{c}=\left|\begin{array}{rrr} 2 & -3 & 4 \\ 1 & 2 & -3 \\ 3 & 4 & -1 \end{array}\right|$$
= 2 (- 2 + 12) + 3 (- 1 + 9) + 4 (4 – 6)
= 20 + 24 – 8 = 36

and $$\vec{a} \cdot(\vec{b} \times \vec{c})=(\vec{b} \times \vec{c}) \cdot \vec{a}$$
= $$\left|\begin{array}{rrr} 1 & 2 & -3 \\ 3 & 4 & -1 \\ 2 & -3 & 4 \end{array}\right|$$
= 1 (16 – 3) – 2 (12 + 2) – 3 (- 9 – 8)
= 13 – 28 + 51 = 36
Thus $$(\vec{a} \times \vec{b}) \cdot \vec{c}=\vec{a} \cdot(\vec{b} \times \vec{c})$$.

Question 2.
If $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$$, $$\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}$$ and $$\vec{c}=3 \hat{i}+4 \hat{j}-\hat{k}$$ then find $$(\vec{a} \times \vec{b}) \cdot \vec{c}$$ and $$\vec{a} \cdot(\vec{b} \times \vec{c})$$. Is $$(\vec{a} \times \vec{b}) \cdot \vec{c}=\vec{a} \cdot(\vec{b} \times \vec{c})$$?
Solution:
Given $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$$,
$$\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}$$
and $$\vec{c}=3 \hat{i}+4 \hat{j}-\hat{k}$$

∴ $$(\vec{a} \times \vec{b})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 4 \\ 1 & 2 & -3 \end{array}\right|$$
= $$\hat{i}(9-8)-\hat{j}(-6-4)+\hat{k}(4+3)$$
= $$\hat{i}+10 \hat{j}+7 \hat{k}$$

Thus $$(\vec{a} \times \vec{b}) \cdot \vec{c}=(\hat{i}+10 \hat{j}+7 \hat{k}) \cdot(3 \hat{i}+4 \hat{j}-\hat{k})$$
= + 1 (3) + 10 (4) + 7 (- 1)
= 36

and $$\vec{b} \times \vec{c}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 3 & 4 & -1 \end{array}\right|$$
= $$\hat{i}(-2+12)-\hat{j}(-1+9)+\hat{k}(4-6)$$
= $$10 \hat{i}-8 \hat{j}-2 \hat{k}$$

∴ $$\vec{a} \cdot(\vec{b} \times \vec{c})=(2 \hat{i}-3 \hat{j}+4 \hat{k}) \cdot(10 \hat{i}-8 \hat{j}-2 \hat{k})$$
= 2 (10) – 3 (- 8) + 4 (- 2)
= 20 + 24 – 8 = 36
Thus, $$(\vec{a} \times \vec{b}) \cdot \vec{c}=\vec{a} \cdot(\vec{b} \times \vec{c})$$.

Question 3.
If $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$$, $$\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$$ and $$\vec{c}=3 \hat{i}-\hat{j}+2 \hat{k}$$, then find
(i) $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]$$
(ii) $$[\vec{a}+\vec{b} \vec{b}+\vec{c} \vec{c}+\vec{a}]$$
Solution:
(i) Given $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$$,
$$\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$$
and $$\vec{c}=3 \hat{i}-\hat{j}+2 \hat{k}$$
$$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{rrr} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{array}\right|$$
= 2 (4 – 1) + 3 (2 + 3) + 4 (- 1 – 6)
= 6 + 15 – 28 = – 7

(ii) Given $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$$,
$$\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$$
and $$\vec{c}=3 \hat{i}-\hat{j}+2 \hat{k}$$

Question 4.
Prove the following :
(i) $$\vec{\alpha} \cdot(\vec{\beta} \times \vec{\alpha})$$ = 0 where $$\vec{\alpha}, \vec{\beta}$$ are any vectors.
(ii) $$[\vec{a}, \vec{b}, \vec{c}+\vec{d}]=[\vec{a}, \vec{b}, \vec{c}]+[\vec{a}, \vec{b}, \vec{d}]$$
Solution:
(i)

Aliter:

(ii) $$[\vec{a}, \vec{b}, \vec{c}+\vec{d}]=\vec{a} \cdot[\vec{b} \times(\vec{c}+\vec{d})]$$
= $$\vec{a} \cdot\{\vec{b} \times \vec{c}+\vec{b} \times \vec{d}\}$$
⇒ $$\vec{a} \cdot(\vec{b} \times \vec{c})+\vec{a} \cdot(\vec{b} \times \vec{d})=\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]+\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{d} \end{array}\right]$$

Question 5.
If $$\hat{i}, \hat{j} \text { and } \hat{k}$$ are three mutually perpendicular unit vectors, then prove that $$\hat{i} \cdot(\hat{k} \times \hat{j})=\hat{j} \cdot(\hat{i} \times \hat{k})=\hat{k} \cdot(\hat{j} \times \hat{i})$$ = – 1.
Solution:
Since $$\hat{i}, \hat{j} \text { and } \hat{k}$$ are mutually ⊥unit vectors

Question 5 (old).
(ii) $$\hat{i}+\hat{j}+\hat{k}, \hat{i}-\hat{i}+\hat{k} \text { and } \hat{i}+2 \hat{j}-\hat{k}$$ (ISC 2003)
Solution:
Given $$\vec{a}=\hat{i}+\hat{j}+\hat{k}$$ ;
$$\vec{b}=\hat{i}-\hat{j}+\hat{k}$$
and $$\vec{c}=\hat{i}+2 \hat{j}-\hat{k}$$
∴ $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{rrr} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{array}\right|$$
= 1 (1 – 2) – 1 (- 1 – 1) + 1 (2 + 1)
= – 1 + 2 + 3
= 4 cubic units
∴ required volume of || piped whose coterminous represented by $$\vec{a}, \vec{b} \text { and } \vec{c}$$
= $$\left|\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]\right|$$
= 4 cubic units

Question 6.
Find the volume of the parallelopiped whose coterminus edges are represented by vectors
(i) $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$$, $$\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$$ and $$\vec{c}=2 \hat{i}-\hat{j}+2 \hat{k}$$
(ii) $$5 \hat{i}-2 \hat{j}+3 \hat{k}$$, $$2 \hat{i}+\hat{j}-\hat{k}$$ and $$3 \hat{i}-\hat{j}+2 \hat{k}$$ (ISC 2003)
Solution:
(i) Let $$\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$$ ;
$$\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$$ ;
$$\vec{c}=2 \hat{i}-\hat{j}+2 \hat{k}$$
∴ Volume of parallelopiped = $$\left|\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]\right|$$
Here, $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{rrr} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 2 & -1 & 2 \end{array}\right|$$;
expanding along R1
= 2 (4 – 1) + 3 (2 + 1) + 4 (- 1 – 4)
= 6 + 12 – 20 = – 2
∴ Volume of parallelopiped = $$\left|\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]\right|$$
= |- 2| = 2 cubic units

(ii) Let $$\vec{a}=5 \hat{i}-2 \hat{j}+3 \hat{k}$$ ;
$$\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$$ ;
$$\vec{c}=3 \hat{i}-\hat{j}+2 \hat{k}$$
∴ $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{rrr} 5 & -2 & 3 \\ 2 & 1 & -1 \\ 3 & -1 & 2 \end{array}\right|$$
= 5 (2 – 1) + 2 (4 + 3) + 3 (- 2- 3)
= 5 + 14 – 15
= 4 cubic units

Question 6 (old).
Show that each of the following triads of vectors are coplanar :
(i) $$\vec{a}=-4 \hat{i}-6 \hat{j}-2 \hat{k}$$, $$\vec{b}=-\hat{i}+4 \hat{j}+3 \hat{k}$$ and $$\vec{c}=-8 \hat{i}-\hat{j}+3 \hat{k}$$ ; $$\hat{i}-2 \hat{j}+3 \hat{k}$$
(ii) $$-2 \hat{i}+3 \hat{j}-4 \hat{k}$$ and $$\hat{i}-3 \hat{j}+5 \hat{k}$$
Solution:
(i) Given $$\vec{a}=-4 \hat{i}-6 \hat{j}-2 \hat{k}$$,
$$\vec{b}=-\hat{i}+4 \hat{j}+3 \hat{k}$$
and $$\vec{c}=-8 \hat{i}-\hat{j}+3 \hat{k}$$
$$\vec{a}, \vec{b}, \vec{c}$$ are coplanar iff their scalar triple product is 0
i.e., $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]$$ = 0
Here $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{rrr} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{array}\right|$$
= – 4 (12 + 3) + 6 (- 3 + 24) – 2 (1 + 32)
= – 60 + 126 – 66 = 0
Thus given vectors are coplanar.

(ii) Given $$\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}$$ ;
$$\vec{b}=-2 \hat{i}+3 \hat{j}-4 \hat{k}$$
and $$\vec{c}=\hat{i}-3 \hat{j}+5 \hat{k}$$
$$\vec{a}, \vec{b}, \vec{c}$$ are coplanar iff their scalar triple product is 0
i.e. $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]$$ = 0
Here $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{rrr} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{array}\right|$$
= 1 (15 – 12) + 2 (- 10 + 4) + 3 (6 – 3)
= 3 – 12 + 9 = 0.

Question 7.
Show that each of the following triads of vectors are coplanar :
(i) $$\vec{a}=-4 \hat{i}-6 \hat{j}-2 \hat{k}$$, $$\vec{b}=-\hat{i}+4 \hat{j}+3 \hat{k}$$ and $$\vec{c}=-8 \hat{i}-\hat{j}+3 \hat{k}$$
(ii) $$\hat{i}-2 \hat{j}+3 \hat{k}$$, $$-2 \hat{i}+3 \hat{j}-4 \hat{k}$$ and $$\hat{i}-3 \hat{j}+5 \hat{k}$$
Solution:
(i) Given $$\vec{a}=-4 \hat{i}-6 \hat{j}-2 \hat{k}$$,
$$\vec{b}=-\hat{i}+4 \hat{j}+3 \hat{k}$$
and $$\vec{c}=-8 \hat{i}-\hat{j}+3 \hat{k}$$
$$\vec{a}, \vec{b}, \vec{c}$$ are coplanar iff their scalar triple product is 0
i.e. $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]$$ = 0

Here $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{rrr} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{array}\right|$$
= – 4 (12 + 3) + 6 (- 3 + 24) – 2 (1 + 32)
= – 60 + 126 – 66 = 0
Thus given vectors are coplanar.

(ii) Given $$\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}$$,
$$\vec{b}=-2 \hat{i}+3 \hat{j}-4 \hat{k}$$
and $$\vec{c}=\hat{i}-3 \hat{j}+5 \hat{k}$$
$$\vec{a}, \vec{b}, \vec{c}$$ are coplanar iff their scalar triple product is 0
i.e. $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]$$ = 0
Here $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]=\left|\begin{array}{rrr} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{array}\right|$$
= 1 (15 – 12) + 2 (- 10 + 4) + 3 (6 – 3)
= 3 – 12 + 9 = 0

Question 7 (old).
Find the value of λ so that the following vectors are coplanar ;
(i) $$\hat{i}+3 \hat{j}$$, $$5 \hat{k}$$ and $$\lambda \hat{i}-\hat{j}$$ (ISC 2006)
Solution:
Given $$\vec{a}=\hat{i}+3 \hat{j}$$ ;
$$\vec{b}=5 \hat{k}$$
and $$\vec{c}=\lambda \hat{i}-\hat{j}$$
Since given vectors $$\vec{a}, \vec{b} \text { and } \vec{c}$$ are coplanar
∴ $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]$$ = 0
⇒ $$\left|\begin{array}{rrr} 1 & 3 & 0 \\ 0 & 0 & 5 \\ \lambda & -1 & 0 \end{array}\right|$$ = 0
⇒ 1 (0 + 5) – 3 (0 – 5λ) + 0 = 0
⇒ 5 + 15λ = 0
⇒ λ = – $$\frac{1}{3}$$

Question 8.
Find the value of λ so that the following vectors are coplanar ;
(i) $$\hat{i}-\hat{j}+\hat{k}$$, $$2 \hat{i}+\hat{j}-\hat{k}$$, $$\lambda \hat{i}-\hat{j}+\lambda \hat{k}$$
(ii) $$2 \hat{i}-\hat{j}+\hat{k}$$, $$\hat{i}+2 \hat{j}-3 \hat{k}$$ and $$3 \hat{i}+\lambda \hat{j}+5 \hat{k}$$
Solution:
(i) Given $$\vec{a}=\hat{i}-\hat{j}+\hat{k}$$ ;
$$\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$$
and $$\vec{c}=\lambda \hat{i}-\hat{j}+\lambda \hat{k}$$
Now given vectors $$\vec{a}, \vec{b} \text { and } \vec{c}$$ are coplanar
∴ $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]$$ = 0
⇒ $$\left|\begin{array}{rrr} 1 & -1 & 1 \\ 2 & 1 & -1 \\ \lambda & -1 & \lambda \end{array}\right|$$ = 0
⇒ 1 (λ – 1) + 1 (2λ + λ) + 1 (- 2 – λ) = 0
⇒ λ – 1 + 3λ – 2 – λ = 0
⇒ 3λ – 3 = 0
⇒ λ = 1

(ii) Given $$\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$$ ;
$$\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}$$
and $$\vec{c}=3 \hat{i}+\lambda \hat{j}+5 \hat{k}$$
Now given vectors $$\vec{a}, \vec{b} \text { and } \vec{c}$$ are coplanar
∴ $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]$$ = 0
⇒ $$\left|\begin{array}{rrr} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & \lambda & 5 \end{array}\right|$$ = 0
⇒ 2 (10 + 3λ) + 1 (5 + 9) + 1 (λ – 6) = 0
⇒ 20 + 6λ + 14 + λ – 6 = 0
⇒ 7λ = – 28
⇒ λ = – $$\frac{28}{7}$$
= – 4

Question 9.
If $$\vec{a}=\hat{i}+\hat{j}+\hat{k}$$, $$\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$$, $$\vec{c}=x \hat{i}+(x-2) \hat{j}-\hat{k}$$ and the vector $$\vec{c}$$ lies in the plane of $$\vec{a} \text { and } \vec{b}$$, then find the value of x.
Solution:
Given, $$\vec{a}=\hat{i}+\hat{j}+\hat{k}$$ ;
$$\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$$ ;
$$\vec{c}=x \hat{i}+(x-2) \hat{j}-\hat{k}$$
Since $$\vec{c}$$ lies in the plane of $$\vec{a} \text { and } \vec{b}$$
∴ $$\vec{a}, \vec{b} \text { and } \vec{c}$$ are coplanar
∴ $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]$$ = 0 ;
Expanding along R1
= 1 [1 – 2 (x – 2)] – 1 [- 1 – 2x] + 1 [x – 2 + x] = 0
⇒ 1 – 2x + 4 + 1 + 2x + 2x – 2 = 0
⇒ 2x + 4 = 0
⇒ x = – 2

Question 9 (old).
If $$\vec{a}=\hat{i}-\hat{k}$$, $$\vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$$ and $$\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$$, then show that $$\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]$$ is independent of x and y.
Solution:
Given $$\vec{a}=\hat{i}-\hat{k}$$,
$$\vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$$
and $$\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$$
Here,
$$\left|\begin{array}{ccc} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{array}\right|$$ ;
operate C3 → C3 + C1
= $$\left|\begin{array}{ccc} 1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1+x \end{array}\right|$$ ;
Expanding along R1
= 1 $$\left|\begin{array}{cc} 1 & 1 \\ x & 1+x \end{array}\right|$$
= 1 + x – x = 1
which is clearly independent of x and y.

Question 10.
Show that the four points with position vectors $$4 \hat{i}+8 \hat{j}+12 \hat{k}$$, $$2 \hat{i}+4 \hat{j}+6 \hat{k}$$, $$3 \hat{i}+5 \hat{j}+4 \hat{k}$$ and $$5 \hat{i}+8 \hat{j}+5 \hat{k}$$ are coplanar.
Solution:
Let A, B, C and D are four given points whose position vectors are $$4 \hat{i}+8 \hat{j}+12 \hat{k}$$, $$2 \hat{i}+4 \hat{j}+6 \hat{k}$$, $$3 \hat{i}+5 \hat{j}+4 \hat{k}$$ and $$5 \hat{i}+8 \hat{j}+5 \hat{k}$$

if – 2(21 – 0) + 4 (7 + 8) – 6 (0 + 3)
if – 42 + 60 – 18 = 0
if 0 = 0, which is true
Hence A, B, C and D are coplanar.

Question 11.
Show that the four points having positión vectors , ,  and  are not coplanar.
Solution:
Let A, B, C and D be given points
Then these given points are coplanar iff any one of triad of vectors are coplanar.
i.e., $$\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}$$ ; $$\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{CD}}$$ etc

= 10 (70 + 72) + 112 (- 30 – 24) – 4 (- 36 + 28)
= 1420-648 + 32
= 804 ≠ 0
∴ $$\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}} \text { and } \overrightarrow{\mathrm{AD}}$$ are non coplanar.
Thus given points A, B, C, and D are non – coplanar.

Question 12.
(i) Find the value of λ for which the four points A, B, C and D with posiotion vectors $$-\hat{j}-\hat{k}$$, $$4 \hat{i}+5 \hat{j}+\lambda \hat{k}$$, $$3 \hat{i}+9 \hat{j}+4 \hat{k}$$ and $$-4 \hat{i}+4 \hat{j}+4 \hat{k}$$ respectively are coplanar. (ISC 2013)
(ii) Find the value of λ for which the four points A, B, C and D with posiotion vectors $$6 \hat{i}-7 \hat{j}$$, $$16 \hat{i}-19 \hat{j}-4 \hat{k}$$, $$\lambda \hat{j}-6 \hat{k}$$ and $$2 \hat{i}-5 \hat{j}+10 \hat{k}$$ are coplanar. (ISC 2017)
Solution:
(i) Let A,B, C and D be given points such that

⇒ 4 (50 – 25) – 6 (15 + 20) + (λ + 1) (15 + 40) = 0
⇒ 100 – 210 + 55 λ + 55 = 0
⇒ – 55 + 55 λ = 0
⇒ λ = 1.

(ii) Let A, B, C and D are the four points whose position vectors are given as under:
$$6 \hat{i}-7 \hat{j}$$, $$16 \hat{i}-19 \hat{j}-4 \hat{k}$$, $$\lambda \hat{j}-6 \hat{k}$$ and $$2 \hat{i}-5 \hat{j}+10 \hat{k}$$

Expanding along R1
10 [10 (λ + 7) + 12] + 12 [- 60 – 24] – 4 [- 12 + 4 (λ + 7)] = 0
⇒ 10 [10λ + 82] + 12 [- 84] – 4 [4λ + 16] = 0
⇒ 100λ + 820 – 1008 – 16λ – 64 = 0
⇒ 84λ – 252 = 0
⇒ λ = $$\frac{252}{84}$$ = 3.

Question 13.
Find λ such that the four points A (- 1, 4, – 3), B (3, λ, – 5), C (- 3, 8, – 5) and D (- 3, 2, 1) are coplanar.
Solution:
Given position vectors of the four points A, B, C and D are ;

⇒ 4 (16 – 4) – (λ – 4) (- 8 – 4) – 2 (4 + 8) = 0
⇒ 48 – (λ – 4) (- 12) – 24 = 0
⇒ 48 + 12λ – 48 – 24 = 0
⇒ 12λ = 24
⇒ λ = 2.

Question 14.
For any three vectors $$\vec{a}, \vec{b}, \vec{c}$$, show that $$\vec{a}-\vec{b}, \vec{b}-\vec{c} \text { and } \vec{c}-\vec{a}$$ are coplanar. (ISC 2016)
Solution:

Question 14 (old).
If $$\vec{a}, \vec{b}, \vec{c}$$ are three vectors, show that $$(\vec{a}+\vec{b}) \cdot[(\vec{b}+\vec{c}) \times(\vec{c}+\vec{a})]=2[\vec{a} \vec{b} \vec{c}]$$. (ISc 2007)
Solution:

Question 15.
If the vectors $$\vec{a}, \vec{b}, \vec{c}$$ are coplanar, show that $$\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}$$ are also coplanar.
Solution: