12th NCERT Vector Algebra Exercise 10.2 Questions 19
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Question (1)

Compute the magnitude of the following vectors:
$\overrightarrow a = \widehat i + \widehat j + \widehat k$
$\overrightarrow b = 2\widehat i - 7\widehat j - 3\widehat k$
$\overrightarrow c = \frac{1}{{\sqrt 3 }}\widehat i + \frac{1}{{\sqrt 3 }}\widehat j - \frac{1}{{\sqrt 3 }}\widehat k$

Solution

$\overrightarrow a = \widehat i + \widehat j + \widehat k$
$\overrightarrow b = 2\widehat i - 7\widehat j - 3\widehat k$
$\overrightarrow c = \frac{1}{{\sqrt 3 }}\widehat i + \frac{1}{{\sqrt 3 }}\widehat j - \frac{1}{{\sqrt 3 }}\widehat k$
$\left| {\overrightarrow a } \right| = \sqrt {1 + 1 + 1} = 3$ $\left| {\overrightarrow b } \right| = \sqrt {4 + 49 + 9} = \sqrt {62}$ $\left| {\overrightarrow c } \right| = \sqrt {\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = 1$

Question (2)

Two different vectors having same magnitude

Solution

$\overrightarrow a = 2\widehat i + \widehat j - \widehat k$ $\overrightarrow b = \widehat i + 2\widehat j + \widehat k$

Question (3)

Two vectors havings same direction

Solution

$\overrightarrow a = 2\widehat i + \widehat j - 3\widehat k$ $\overrightarrow b = 4\widehat i + 2\widehat j - 6\widehat k$

Question (4)

Find the values of x and y so that the vectors $\overrightarrow a = 2\widehat i + 3\widehat j$ and $\overrightarrow b = x\hat i + y\hat j$ are equal

Solution

$\overrightarrow a = 2\widehat i + 3\widehat j$ $\overrightarrow b = x\hat i + y\hat j$ Vectors are equal $\overrightarrow a = \overrightarrow b$ $2\widehat i + 3\widehat j = x\hat i + y\hat j$ $x = 2,\quad y = 3$

Question (5)

Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (-5, 7).

Solution

$A\left( {2,1} \right)\qquad B\left( { - 5,7} \right)$ $\overrightarrow {AB} = \left( { - 5 - 2} \right)\widehat i + \left( {7 - 1} \right)\widehat j$ $\overrightarrow {AB} = - 7\widehat i + 6\widehat j$

Question (6)

Find the sum of vectrs $\overrightarrow a = \widehat i - 2\widehat j + \widehat k$, $\overrightarrow b = - 2\widehat i + 4\widehat j + 5\widehat k$   and $\overrightarrow c = \widehat i - 6\widehat j - 7\widehat k$

Solution

$\overrightarrow a = \widehat i - 2\widehat j + \widehat k$ $\overrightarrow b = - 2\widehat i + 4\widehat j + 5\widehat k$ $\overrightarrow c = \widehat i - 6\widehat j - 7\widehat k$ $\overrightarrow a + \overrightarrow b + \overrightarrow c = 0\widehat i - 4\widehat j - \widehat k$

Question (7)

Find the unit vector in the direction of vector $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$

Solution

$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$ $\therefore \left| {\overrightarrow a } \right| = \sqrt {1 + 1 + 4} = \sqrt 6$ $\text{Unit vector}\quad \widehat a = \frac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}}$ $\widehat a = \frac{1}{{\sqrt 6 }}\widehat i + \frac{1}{{\sqrt 6 }}\widehat j + \frac{1}{{\sqrt 6 }}\widehat k$

Question (8)

Find the unit vector in in the direction of vector $\overrightarrow {PQ}$, where P and Q are the points (1,2,3) and (4, 5, 6) respectively

Solution

$P\left( {1,2,3} \right)\qquad Q\left( {4,5,6} \right)$ $\overrightarrow {PQ} = 3\widehat i + 3\widehat j + 3\widehat k$ $\left| {\overrightarrow {PQ} } \right| = \sqrt {9 + 9 + 9} = \sqrt {27} = 3\sqrt 3$ $\widehat {PQ} = \frac{{\overrightarrow {PQ} }}{{\left| {\overrightarrow {PQ} } \right|}}$ $\widehat {PQ} = \frac{{3\widehat i + 3\widehat j + 3\widehat k}}{{3\sqrt 3 }}$ $\widehat {PQ} = \frac{1}{{3\sqrt 3 }}\widehat i + \frac{1}{{3\sqrt 3 }}\widehat j + \frac{1}{{3\sqrt 3 }}\widehat k$

Question (9)

For given vector $\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k$ and $\overrightarrow b = - \widehat i + \widehat j - \widehat k$, find the unit vector in the direction of the vector $\overrightarrow a + \overrightarrow b$

Solution

$\overrightarrow a = 2\widehat i - \widehat j + 2\widehat k$ $\overrightarrow b = - \widehat i + \widehat j - \widehat k$ $\overrightarrow a + \overrightarrow b = 2\widehat i - \widehat j + 2\widehat k - \widehat i + \widehat j - \widehat k$ $\overrightarrow a + \overrightarrow b = \widehat i + \widehat k$ $\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {1 + 1} = \sqrt 2$ $\frac{{\overrightarrow a + \overrightarrow b }}{{\left| {\overrightarrow a + \overrightarrow b } \right|}} = \frac{{\widehat i + \widehat j}}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\widehat i + \frac{1}{{\sqrt 2 }}\widehat k$

Question (10)

Find a vector in the direction of vector $\overrightarrow a = 5\widehat i - \widehat j + 2\widehat k$, which has magnitude 8 units

Solution

$\overrightarrow a = 5\widehat i - \widehat j + 2\widehat k$ $\left| {\overrightarrow a } \right| = \sqrt {25 + 1 + 4} = \sqrt {30}$ $\text{vector of magnitude 8} = 8\frac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}}$ $= 8\frac{{\left( {5\widehat i - \widehat j + 2\widehat k} \right)}}{{\sqrt {30} }}$ $= \frac{{40}}{{\sqrt {30} }}\widehat i - \frac{8}{{\sqrt {30} }}\widehat j + \frac{{16}}{{\sqrt {30} }}\widehat k$

Question (11)

Show that $\overrightarrow a = 2\widehat i - 3\widehat j + 4\widehat k$ and $\overrightarrow b = - 4\widehat i + 6\widehat j - 8\widehat k$ are collinear

Solution

$\overrightarrow a = 2\widehat i - 3\widehat j + 4\widehat k$ $\overrightarrow b = - 4\widehat i + 6\widehat j - 8\widehat k$ $\overrightarrow a \times \overrightarrow b = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\2&{ - 3}&4\\{ - 4}&6&{ - 8}\end{array}} \right|$ $\overrightarrow a \times \overrightarrow b = \left( {24 - 24} \right)\widehat i - \left( { - 16 + 16} \right)\widehat j + \left( {12 - 12} \right)\widehat k$ $\overrightarrow a \times \overrightarrow b = \overrightarrow 0$ ⇒ $\overrightarrow a$ and $\overrightarrow b$ are both collinear

Question (12)

Find the direction cosine of the vector $\overrightarrow a = \widehat i + 2\widehat j + 3\widehat k$

Solution

$\overrightarrow a = \widehat i + 2\widehat j + 3\widehat k$ $\left| {\overrightarrow a } \right| = \sqrt {1 + 4 + 9} = \sqrt {14}$ $\widehat a = \frac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}}$ $\widehat a = \frac{{\widehat i + 2\widehat j + 3\widehat k}}{{\sqrt {14} }}$ $\text{cosine direction} = \left( {\frac{1}{{\sqrt {14} }},\frac{2}{{\sqrt {14} }},\frac{3}{{\sqrt {14} }}} \right)$

Question (13)

Find the direction cosines of the vector joining the points A(1, 2, -3) and B(-1, -2, 1), directed from A to B

Solution

$A\left( {1,2, - 3} \right),\text{and} \quad B\left( { - 1, - 2,1} \right)$ $\overrightarrow {AB} = \left( { - 1, - 2,1} \right) - \left( {1,2, - 3} \right)$ $\overrightarrow {AB} = \left( { - 2, - 4,4} \right)$ $\overrightarrow {AB} = - 2\widehat i - 4\widehat j + 4\widehat k$ $\left| {\overrightarrow {AB} } \right| = \sqrt {4 + 16 + 16} = \sqrt {36} = 6$ $\text{Direction cosine} = \frac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|}}$ $\text{Direction cosine} = \frac{{ - 2\widehat i - 4\widehat j + 4\widehat k}}{6}$ $\text{Direction cosine} = \frac{{ - 1}}{3}\widehat i - \frac{2}{3}\widehat j + \frac{2}{3}\widehat k$

Question (14)

Show that vector $\overrightarrow a = \widehat i + \widehat j + \widehat k$ is equally iinclined to the axes OX, OY and OZ

Solution

$\overrightarrow a = \widehat i + \widehat j + \widehat k$ $\left| {\overrightarrow a } \right| = \sqrt {1 + 1 + 1} = \sqrt 3$ $\text{cosine direction} = \frac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}}$ $\text{cosine direction} = \frac{{\widehat i + \widehat j + \widehat k}}{{\sqrt 3 }}$ $\left( {\cos \alpha ,\cos \beta ,\cos \delta } \right) = \left( {\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)$ $\cos \alpha = \frac{1}{{\sqrt 3 }}$ $\alpha = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$ $\beta = \gamma = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$ $\text{as } \quad \alpha = \beta = \gamma$ so it is equally inclined

Question (15)

Find the position vector of a point R which divides the line joining two points P and Q whose position vector of a point R which divides the line joining two points P and Q whose position vectors are $\overrightarrow p = \widehat i + 2\widehat j - \widehat k\quad and \quad \overrightarrow q = - \widehat i + \widehat j + \widehat k$ respectively, in the ratio 2:1
(i) internally and (ii) externally

Solution

$\overrightarrow p = \widehat i + 2\widehat j - \widehat k\quad and \quad \overrightarrow q = - \widehat i + \widehat j + \widehat k$ (i) internally
$R\left( {\overrightarrow r } \right)$ divide $\overrightarrow {PQ}$ internally in ratio 2:1
$\therefore \overrightarrow r = \frac{{2\overrightarrow q + \overrightarrow q }}{{2 + 1}}$ $\overrightarrow r = \frac{{2\left( { - \widehat i + \widehat j + \widehat k} \right) + \left( {\widehat i + 2\widehat j - \widehat k} \right)}}{3}$ $\overrightarrow r = \frac{{ - \widehat i + 4\widehat j + \widehat k}}{3}$

(ii) External
R divides PQ externally in ratio 2:1
$\overrightarrow r = \frac{{2\left( { - \widehat i + \widehat j + \widehat k} \right) + \left( {\widehat i + 2\widehat j - \widehat k} \right)}}{{ - 1}}$ $\overrightarrow r = \frac{{3\widehat i + 0\widehat j - 3\widehat k}}{{ - 1}}$ $\overrightarrow r = - 3\widehat i + 3\widehat k$

Question (16)

Find the position vector of the mid point of the vector jpining the points P(2, 3, 4) and Q(4, 1, -2)

Solution

$P\left( {2,3,4} \right),\quad and \quad Q\left( {4,1, - 2} \right)$ Let R is mid point of $\overrightarrow {PQ}$
$\therefore \overrightarrow r = \frac{{\overrightarrow p + \overrightarrow q }}{2}$ $\overrightarrow r = \left( {\frac{{2 + 4}}{2},\frac{{3 + 1}}{2},\frac{{4 - 2}}{2}} \right)$ $\overrightarrow r = \left( {3,2,1} \right)$

Question (17)

Show that the points A, B, and C with position vectors ${\overrightarrow a } = 3\widehat i - 4\widehat j - 4\widehat k$ , ${\overrightarrow b } = 2\widehat i - \widehat j + \widehat k$ and ${\overrightarrow c } = \widehat i - 3\widehat j - 5\widehat k$

Solution

$A\left( {\overrightarrow a } \right) = 3\widehat i - 4\widehat j - 4\widehat k$ $B\left( {\overrightarrow b } \right) = 2\widehat i - \widehat j + \widehat k$ $C\left( {\overrightarrow c } \right) = \widehat i - 3\widehat j - 5\widehat k$ $\overrightarrow {AB} = \overrightarrow b - \overrightarrow a = - \widehat i + 3\widehat j + 5\widehat k$ $AB = \left| {\overrightarrow {AB} } \right| = \sqrt {1 + 9 + 25} = \sqrt {35}$ $\overrightarrow {BC} = \overrightarrow c - \overrightarrow b = - \widehat i - 2\widehat j - 6\widehat k$ $BC = \overrightarrow {BC} = \sqrt {1 + 4 + 36} = \sqrt {41}$ $\overrightarrow {AC} = \overrightarrow c - \overrightarrow a = - 2\widehat i + \widehat j - \widehat k$ $AC = \overrightarrow {AC} = \sqrt {4 + 1 + 1} = \sqrt 6$ $A{B^2} + A{C^2} = 35 + 6$ $A{B^2} + A{C^2} = 41$ $A{B^2} + A{C^2} = B{C^2}$ ⇒ By converse of Pythagorus ∠A =90° is right angle in δABC
$\overrightarrow {AB \cdot } \overrightarrow {AC} = \left( { - \widehat i + 3\widehat j + 5\widehat k} \right) \cdot \left( { - 2\widehat i + \widehat j - \widehat k} \right)$ $\overrightarrow {AB \cdot } \overrightarrow {AC} = 2 + 3 - 5 = 0$ $\Rightarrow \overrightarrow {AB} \bot \overrightarrow {AC}$ $\Rightarrow \angle A = 90$

Question (18)

In traiangle ABC (Fig 10.18), which of the following is not true

$(A) \quad \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CA} = \overrightarrow 0$
$(B) \quad\overrightarrow {AB} + \overrightarrow {BC} - \overrightarrow {AC} = \overrightarrow 0$
$(C) \quad \overrightarrow {AB} + \overrightarrow {BC} - \overrightarrow {CA} = \overrightarrow 0$
$(D) \quad \overrightarrow {AB} - \overrightarrow {CB} + \overrightarrow {CA} = \overrightarrow 0$

Solution

$\text{By triangle law}\overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC}$ $\overrightarrow {AB} + \overrightarrow {BC} - \overrightarrow {AC} = 0$ $\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CA} = 0$ not correct

Question (19)

If &\overrightarrow a & and &\overrightarrow b & are two collinear vectors, then which of the following are incorrect
$(A) \ quad \overrightarrow b = \lambda \overrightarrow a$, for some scalar λ
$(B) \quad \overrightarrow a = \pm \overrightarrow b$
(C) the respective components of &\overrightarrow a & and &\overrightarrow b & are not proportional
(D) both the vectors &\overrightarrow a & and &\overrightarrow b & have same direction, but different magniitudes

Solution

$\overrightarrow a \quad and \quad \overrightarrow b \text{colliner then}$ $\overrightarrow b = \lambda \overrightarrow a$