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

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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: