Referring to the AP Inter 2nd Year Maths Study Material Chapter 10 Vector Algebra Exercise 10b Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Vector Algebra Solutions Exercise 10b
I.
Question 1.
Compute the magnitude of the vector \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
Solution:
Given that \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
∴ \(|\vec{a}|=\sqrt{x^2+y^2+z^2}=\sqrt{1+1+1}=\sqrt{3}\)
Question 2.
Compute the magnitude of the vector \(\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\)
Solution:
Given that \(\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\)
\(|\overrightarrow{\mathrm{b}}|=\sqrt{(2)^2+(-7)^2+(-3)^2}=\sqrt{4+49+9}=\sqrt{62} .\)
Question 3.
Compute the magnitude of the vector \(\vec{c}=\frac{1}{\sqrt{3}} \hat{\mathbf{i}}+\frac{1}{\sqrt{3}} \hat{\mathbf{j}}-\frac{1}{\sqrt{3}} \hat{\mathbf{k}}\)
Solution:
Given that \(\vec{c}=\frac{1}{\sqrt{3}} \hat{\mathbf{i}}+\frac{1}{\sqrt{3}} \hat{\mathbf{j}}-\frac{1}{\sqrt{3}} \hat{\mathbf{k}}\)
\(|\overrightarrow{\mathrm{c}}|=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{-1}{\sqrt{3}}\right)^2}=\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=\sqrt{\frac{3}{3}}=\sqrt{1}=1 .\)
Question 4.
Write two different vectors having same direction.
Solution:
Let \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\) and \(\vec{b}=2(\hat{i}+2 \hat{j}+3 \hat{k})=2 \vec{a}\)
Then \(\overrightarrow{\mathrm{b}}=\mathrm{m} \overrightarrow{\mathrm{a}}\) where m = 2 > 0.
∴ Vectors \(\vec{a} \text { and } \vec{b}\) have the same direction. But \(\overrightarrow{\mathrm{b}} \neq \overrightarrow{\mathrm{a}}\) as the corresponding components are distinct. We can write an infinite number of such vectors.
Question 5.
Write two different vectors having same magnitude.
Solution:
Let \(\vec{a}=(\hat{i}-2 \hat{j}+3 \hat{k})\) and \(\overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}})\)
\(|\vec{a}|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{1+4+9}=\sqrt{14}\)
\(|\vec{b}|=\sqrt{2^2+1^2+(-3)^2}=\sqrt{4+1+9}=\sqrt{14}\)
But \(\overrightarrow{\mathrm{a}} \neq \overrightarrow{\mathrm{b}}\) as the corresponding components are distinct.
We can write an infinite number of such vectors.
Question 6.
Find the values of x and y so that the vectors \(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} \text { and } x \hat{\mathbf{i}}+y \hat{\mathbf{j}}\) are equal.
Solution:
Given \(2 \hat{i}+3 \hat{j}=x \hat{i}+y \hat{j}\)
Comparing coefficients of \(\hat{i} \text { and } \hat{j}\) on both sides, we have x = 2 and y = 3.
Question 7.
Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (- 5, 7).
Solution:
Let \(\overrightarrow{\mathrm{AB}}\) be the vector with initial point A(2, 1) and terminal point B(-5, 7)
⇒ PV (Position Vector)of point A(2, 1)is \(\overrightarrow{\mathrm{OA}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}\) and P.V. of point B (-5, 7) is \(\overrightarrow{\mathrm{OB}}=-5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = PV of point B – PV of point A = \((-5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}})-(2 \hat{\mathrm{i}}+\hat{\mathrm{j}})=-5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}\)
⇒ \(\overrightarrow{\mathrm{AB}}=-7 \hat{\mathrm{i}}+6 \hat{\mathrm{j}} .\)
∴ By definition, scalar components of the vectors \(\overrightarrow{\mathrm{AB}}\) are -7 and 6 and vector components of the vector \(\overrightarrow{\mathrm{AB}}\) are \(-7 \hat{\mathrm{i}} \text { and } 6 \hat{\mathrm{j}} .\)
Question 8.
Find the sum of the vectors \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, b=-2 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{c}=\hat{\mathrm{i}}-6 \hat{\mathrm{j}}-7 \hat{\mathrm{k}} .\)
Solution:
Given vectors \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k} \text { and } \vec{c}=\hat{i}-6 \hat{j}-7 \hat{k} .\)
∴ \(\vec{a}+\vec{b}+\vec{c}=(1-2+1) \hat{i}+(-2+4-6) \hat{j}+(1+5-7) \hat{k}=0 \hat{i}-4 \hat{j}-\hat{k}=-4 \hat{j}-\hat{k}\)
Question 9.
Find the unit vector in the direction of the vector \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}} .\)
Solution:
We know that a unit vector in the direction of the vector
\(\vec{a}=\hat{i}+\hat{j}+2 \hat{k} \text { is } \hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{\hat{i}+\hat{j}+2 \hat{k}}{\sqrt{1+1+4}} \Rightarrow \hat{a}=\frac{\hat{i}+\hat{j}+2 \hat{k}}{\sqrt{6}}=\frac{1}{\sqrt{6}} \hat{i}+\frac{1}{\sqrt{6}} \hat{j}+\frac{2}{\sqrt{6}} \hat{k} .\)
Question 10.
Find the unit vector in the direction of vector \(\overrightarrow{\mathrm{PQ}}\), where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.
Solution:
Given points are P(1, 2, 3) and Q(4, 5, 6)
∴ Position vector of the point P(1, 2, 3) is \(\overrightarrow{\mathrm{OP}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
and position vector of point Q(4, 5, 6) is \(\overrightarrow{\mathrm{OQ}}=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\) where O is the origin.
∴ \(\overrightarrow{P Q}=\overrightarrow{O Q}-\overrightarrow{O P}=(4 \hat{i}+5 \hat{j}+6 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})=3 \hat{i}+3 \hat{j}+3 \hat{k}\)
∴ a unit vector in the direction of vector \(\overline{\mathrm{PQ}}=\frac{\overline{\mathrm{PQ}}}{|\overline{\mathrm{PQ}}|}=\frac{3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}}{\sqrt{9+9+9=27=9 \times 3}}\)
\(\frac{3(\hat{i}+\hat{j}+\hat{k})}{3 \sqrt{3}}=\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}\)
Question 11.
For given vectors, \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\). Find the unit vector in the direction of \(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} .\)
Solution:
Given vectors \(\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\)
∴ \(\vec{a}+\vec{b}=2 \hat{i}-\hat{j}+2 \hat{k}-\hat{i}+\hat{j}-\hat{k}=\hat{i}+0 \hat{j}+\hat{k}\)
∴ \(|\vec{a}+\vec{b}|=\sqrt{(1)^2+(0)^2+(1)^2}=\sqrt{2}\)
∴ A unit vector in the direction of \(\vec{a}+\vec{b} \text { is } \frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{\hat{i}+0 \hat{j}+\hat{k}}{\sqrt{2}}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k} .\)
Question 12.
Find a vector in the direction of vector \(5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) which has magnitude 8 units.
Solution:
Let \(\vec{a}=5 \hat{i}-\hat{j}+2 \hat{k} .\)
∴ A vector in the direction of vector \(\vec{a}\) which has magnitude 8 units = \(8 \hat{a}=8 \frac{\vec{a}}{|\vec{a}|}=\frac{8(5 \hat{i}-\hat{j}+2 \hat{k})}{\sqrt{25+1+4}}\)
= \(\frac{8}{\sqrt{30}}(5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=\frac{40}{\sqrt{30}} \hat{\mathrm{i}}-\frac{8}{\sqrt{30}} \hat{\mathrm{j}}+\frac{16}{\sqrt{30}} \hat{\mathrm{k}}\)
Question 13.
Show that the vectors \(2 \hat{i}-3 \hat{j}+4 \hat{k} \text { and }-4 \hat{i}+6 \hat{j}-8 \hat{k}\) are coilinear.
Solution:
Let \(\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k} \text { and } \vec{b}=-4 \hat{i}+6 \hat{j}-8 \hat{k}=-2(2 \hat{i}-3 \hat{j}+4 \hat{k})=-2 \vec{a}\)
Here, \(\overrightarrow{\mathrm{b}}=-2 \overrightarrow{\mathrm{a}}=\mathrm{m} \overrightarrow{\mathrm{a}}\) where m = -2 < 0
∴ Vectors \(\vec{a} \text { and } \vec{b}\) are coilinear (unlike because m = -2 < 0)
Question 14.
Find the direction cosines of the vector \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
Solution:
The given vector is \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \Rightarrow|\vec{a}|=\sqrt{i^2+2^2+3^2}=\sqrt{24}\)
\(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{\hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{14}}=\frac{1}{\sqrt{14}} \hat{i}+\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}\)
We know that direction consines of a vectors \(\vec{a}\) are coefficients of \(\hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}} \text { in } \hat{\mathrm{a}} \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\)
∴ Dc’s of the given vector = \(\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)\)
Question 15.
Find the direction c„osines of the vector joining the points A (1, 2, -3) and B (-1, -2, 1), directed from A to B.
Solution:
Given points A (1, 2, -3) andB (-1, -2, 1) ⇒ \(\overrightarrow{O A}=\hat{i}+2 \hat{j}-3 \hat{k}, \overrightarrow{O B}=-\hat{i}-2 \hat{j}+\hat{k}\)
\(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=(-1-1) \hat{i}+(-2-2) \hat{j}+[1-(-3)] \hat{k}=-2 \hat{i}-4 \hat{j}+4 k\)
\(|\overrightarrow{\mathrm{AB}}|=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{4+16+16}=\sqrt{36}=6\)
∴ A unit vector along AB = \(\frac{\overline{A B}}{|\overrightarrow{A B}|}=\frac{-2 \hat{i}-4 \hat{j}+4 \hat{k}}{6}=-\frac{2}{6} \hat{i}-\frac{4}{6} \hat{j}+\frac{4}{6} \hat{k}=\frac{-1}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k} .\)
∴ Direction Cosines of the vector \(\frac{-1}{3}, \frac{-2}{3}, \frac{2}{3}\)
Question 16.
Show that the vector \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) is equally inclined to the axes OX, OY and OZ.
Solution:
Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\)
\(|\overrightarrow{\mathbf{a}}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}\)
Thus, the DCs of \(\overrightarrow{\mathrm{a}} \text { are }\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\)
Now, α β and γ be the angles formed by \(\vec{a}\) with the positive directions of x, y and z axes respectively.
Then, cosα = \(\frac{1}{\sqrt{3}}\), cosβ = \(\frac{1}{\sqrt{3}}\), cosγ = \(\frac{1}{\sqrt{3}}\)
Hence, the vector is equally inclined to OX, OY and OZ.
Question 17.
Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, -2).
Solution:
The position vector of the midpoint R is \(\overrightarrow{\mathrm{OR}}=\frac{\overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{OQ}}}{2}=\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+(4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{2}\)
\(=\frac{(2+4) \hat{i}+(3+1) \hat{j}+(4-2) \hat{k}}{2}=\frac{6 \hat{i}+4 \hat{j}+2 \hat{k}}{2}=3 \hat{i}+2 \hat{j}+\hat{k}\)
II.
Question 1.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \text { and }-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) respectively in the ratio 2 : 1.
(i) internally
(ii) externally
Solution:
Position vectors of P and Q are given as \(\overrightarrow{\mathrm{OP}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{OQ}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
i) The position vector of R which divides the line joining two points P and Q internally in the ratio
2 : 1 is \(\overline{O R}=\frac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2 \hat{j}-\hat{k})}{2+1}=\frac{-2 \hat{i}+2 \hat{j}+2 \hat{k}+\hat{i}+2 \hat{j}-\hat{k}}{3}\)
= \(\frac{-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{3}=\frac{-1}{3} \hat{\mathrm{i}}+\frac{4}{3} \hat{\mathrm{j}}+\frac{1}{3} \hat{\mathrm{k}}\)
ii) The position vector of R which divides the line joining two points P and Q externally in the 2 : 1 is \(\overrightarrow{\mathrm{OR}}=\frac{2(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})-1(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})}{2-1}=\frac{-2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}-\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{1}=-3 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}\)
Question 2.
Show that the points A, B and C with position vectors, \(\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}, \quad \vec{b}=2 \hat{i}-\hat{j}+\hat{k}\) and \(\vec{c}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\) respectively form the vertices of a right angled triangle.
Solution:
Let \(\overline{\mathrm{OA}}=3 \overline{\mathrm{i}}-4 \overline{\mathrm{j}}-4 \overline{\mathrm{k}}, \overline{\mathrm{OB}}=2 \overline{\mathrm{i}}-\overline{\mathrm{j}}+\overline{\mathrm{k}}, \overline{\mathrm{OC}}=\overline{\mathrm{i}}-3 \overline{\mathrm{j}}-5 \overline{\mathrm{k}}\)
\(\overline{A B}=\overline{O B}-\overline{O A}=(2 \bar{i}-\bar{j}+\bar{k})-(3 \bar{i}-4 \bar{j}-4 \bar{k})=-\bar{i}+3 \bar{j}+5 \bar{k} \Rightarrow|\overline{A B}|=\sqrt{1+9+25}=\sqrt{35}\)
\(\overline{\mathrm{BC}}=\overline{\mathrm{OC}}-\overline{\mathrm{OB}}=(\overline{\mathrm{i}}-3 \overline{\mathrm{j}}-5 \overline{\mathrm{k}})-(2 \overline{\mathrm{i}}-\overline{\mathrm{j}}+\overline{\mathrm{k}})=-\overline{\mathrm{i}}-2 \overline{\mathrm{j}}-6 \overline{\mathrm{k}} \Rightarrow|\overline{\mathrm{BC}}|=\sqrt{1+4+36}=\sqrt{41}\)
\(\overline{C A}=\overline{O A}-\overline{O C}=(3 \bar{i}-4 \bar{j}-4 \bar{k})-(\bar{i}-3 \bar{j}-5 \bar{k})=2 \bar{i}-\bar{j}+\bar{k} \Rightarrow|\overline{C A}|=\sqrt{2+1+1}=\sqrt{6}\)
Here \(|\overrightarrow{\mathrm{OC}}|=(\sqrt{41})^2=41=35+6=|\overrightarrow{\mathrm{AB}}|^2+|\overrightarrow{\mathrm{CA}}|^2\)
∴ Points A, B, C are the vertices of a right angled triangle.