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. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-5/
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: