These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.5 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.5
Question 1.
Find [a, b, c] if \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\), \(\vec{b}=2 \hat{i}-3 \hat{j}+\hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}-2 \hat{k}\)
Solution:
\([\vec{a}, \vec{b}, \vec{c}]=\left|\begin{array}{ccc} 1 & -2 & 3 \\ 2 & -3 & 1 \\ 3 & 1 & -2 \end{array}\right|\) = 1(6 – 1)+2(- 4 – 3) + 3(2 + 9) = 1(5) + 2(- 7) + 3(11) = 24.
Question 2.
Show that the vectors \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\), \(\vec{b}=-2 \hat{i}+3 \hat{j}-4\hat{k}\) and \(\vec{c}=\hat{i}-3\hat{j}+5 \hat{k}\) are coplanar.
Solution:
\([\vec{a}, \vec{b}, \vec{c}]=\left|\begin{array}{ccc} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{array}\right|\)
= 1(15 – 12)+2(- 10 + 4) + 3(6 – 3)
= 1(3) + 2(- 6) + 3(3) = 0.
Question 3.
Find λ, if the vectors \(\hat{i}-\hat{j}+\hat{k}, 3 \hat{i}+\hat{j}+2 \hat{k}\) and \(\hat{i}+\lambda \hat{j}-3 \hat{k}\)are coplanar.
Solution:
Let \(\vec{a}=\hat{i}-\hat{j}+\hat{k}\), \(\vec{b}=3 \hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{c}=\hat{i}+\lambda \hat{j}-3 \hat{k}\) be the given vectors.
Since the given vectors are coplanar \([\vec{a}, \vec{b}, \vec{c}]\) = 0
i.e., \(\left|\begin{array}{ccc} 1 & -1 & 1 \\ 3 & 1 & 2 \\ 1 & \lambda & -3 \end{array}\right|\) = 0
1(- 3 – 2λ) + 1(- 9 – 2) + 1(3λ – 1) = 0
– 3 – 2λ – 11 + 3λ – 1 = 0 ⇒ λ = 15
Question 4.
i. If c1 = 1 and c2 = 2, find c3 which makes \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) coplanar.
ii. Then if c2 = – 1 and c3 = 1, show that no value of c1 can make \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) coplanar.
Solution:
i. Given that \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar.
∴ \([\vec{a}, \vec{b}, \vec{c}]\) = 0
Question 5.
Consider the 4 points A, B, C and D with position vectors \(4 \hat{i}+8 \hat{j}+12 \hat{k}, 2 \hat{i}+4 \hat{j}+6 \hat{k}\) \(3 \hat{i}+5 \hat{j}+4 \hat{k} \text { and } 5 \hat{i}+8 \hat{j}+5 \hat{k}\).
i. Find \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}\).
ii. Show that the 4 points are coplanar.
Solution:
Question 6.
Find x such that the four points A(3, 2, 1), B(4, x, 5), C(4, 2, – 2) and D(6, 5, – 1) are coplanar
Solution:
ii. Since the 4 points A, B, C and D are coplanar \([\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}]\) = 0
\(\left|\begin{array}{ccc}
1 & (x-2) & 4 \\
1 & 0 & -3 \\
3 & 3 & -2
\end{array}\right|\) = 0
1(0 + 9) – (x – 2)(- 2+9) + 4(3 – 0) = 0
9 – 7(x – 2) + 12 = 0
9 – 7x + 14 + 12 = 0
⇒ 7x = 35
∴ x = 5
Question 7.
Show that the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar if \(\vec{a}\) + \(\vec{b}\), \(\vec{b}\) + \(\vec{c}\), \(\vec{c}\) + \(\vec{a}\) are coplanar.
Solution: