12th NCERT Vector Algebra Miscellineous Questions 19
Do or do not
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Question (1)

Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.

Solution

Vector is in xy-plane. Let vector makes angle α with x axis and β with y-axis

α = 30°, β = 60°
Vector $\overrightarrow a = \left( {\cos \alpha ,\cos \beta } \right)$
$\overrightarrow a = \left( {\cos 30,\cos 60} \right)$
$\overrightarrow a = \left( {\frac{{\sqrt 3 }}{2},\frac{1}{2}} \right)$
$\left| {\overrightarrow a } \right| = 1$
$ \therefore \overrightarrow a = \frac{{\sqrt 3 }}{2}\widehat i + \frac{1}{2}\widehat j$

Question (2)

Find the scalar components and magnitude of the vector joining the points P(x1, y1, z1) and Q(x2, y2, z2)

Solution

P( x1, y1, z1) Q( x2, y2, z2)
Scalar component of PQ
$\overrightarrow {PQ} = \left( {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right)$
magnitude of PQ =$\left| {\overrightarrow {PQ} } \right|$
$\left| {\overrightarrow {PQ} } \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $

Question (3)

A girl walks 4km towards west, then she walks 3km in a direction 30° east of north and stops. Determine the girl's displacement from her initital point of departure.

Solution


Let girl start from point O is origine goes towards west 4km
∴ B = (-4, 0)
$\overrightarrow {OB} = - 4\widehat i$
Then she walks 3km towards east of north by 30°
∴ α = 30°, θ = 60°
$\overrightarrow {BC} = \left( {3\cos 60,3\sin 60} \right)$
$\overrightarrow {BC} = \left( {\frac{3}{2},\frac{{3\sqrt 3 }}{2}} \right)$
$\overrightarrow {BC} = \frac{3}{2}\widehat i + \frac{{3\sqrt 3 }}{2}\widehat j$
$\overrightarrow {OC} = \overrightarrow {OB} + \overrightarrow {BC} $
$\overrightarrow {OC} = - 4\widehat i + \frac{3}{2}\widehat i + \frac{{3\sqrt 3 }}{2}\widehat j$
$\overrightarrow {OC} = - \frac{5}{2}\widehat i + \frac{{3\sqrt 3 }}{2}\widehat j$

Question (4)

If $\overrightarrow a = \overrightarrow b + \overrightarrow c $, then is it true that$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| + \left| {\overrightarrow c } \right|$? Justfy your answer

Solution

If $\overrightarrow a = \overrightarrow b + \overrightarrow c $
then $\left| {\overrightarrow a } \right| \ne \left| {\overrightarrow b } \right| + \left| {\overrightarrow c } \right|$
Let $\overrightarrow b = 2\widehat i + 3\widehat j - \widehat k$
$ \therefore \left| {\overrightarrow b } \right| = \sqrt {4 + 9 + 1} = \sqrt {14} $
$\overrightarrow a = \overrightarrow b + \overrightarrow c $
$\overrightarrow a = 5\widehat i + 2\widehat j + \widehat k$
$\left| {\overrightarrow a } \right| = \sqrt {25 + 4 + 1} = \sqrt {30} $
$RHS = \left| {\overrightarrow b } \right| + \left| {\overrightarrow c } \right|$
$RHS = \sqrt {14} + \sqrt {14} = 2\sqrt {14} $ $RHS \ne \sqrt {30} \ne \left| {\overrightarrow a } \right|$

Question (5)

Find the value of x for which $x\left( {\widehat i + \widehat j + \widehat k} \right)$ is a unit vector

Solution

$\overrightarrow a = x\left( {\widehat i + \widehat j + \widehat k} \right)$ is unit vector
$\overrightarrow a = x\widehat i + x\widehat j + x\widehat k$
$\left| {\overrightarrow a } \right| = 1$
$\therefore \sqrt {{x^2} + {x^2} + {x^2}} = 1$
$\sqrt {3{x^2}} = 1$
${x^2} = \frac{1}{3}$
$x = \pm \frac{1}{{\sqrt 3 }}$

Question (6)

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k$ and $\overrightarrow b = \widehat i - 2\widehat j + \widehat k$

Solution

$\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k$ and $\overrightarrow b = \widehat i - 2\widehat j + \widehat k$ Resultant vector $ = \overrightarrow a + \overrightarrow b = 3\widehat i + \widehat j$
Unit vector $ = \frac{{3\widehat i + \widehat j}}{{\sqrt {9 + 1} }}$
Unit vector $ = \frac{{3\widehat i + \widehat j}}{{\sqrt {10} }}$
Vector is of magnitude 5 in direction
$ = 5\left( {\frac{{3\widehat i + \widehat j}}{{\sqrt {10} }}} \right) = \frac{{\cancel{5}\sqrt {10} }}{{ \require{cancel} \cancel{10}_2}}\left( {3\widehat i + \widehat j} \right)$
$ = \frac{{3\sqrt {10} }}{2}\widehat i + \frac{{\sqrt {10} }}{2}\widehat j$

Question (7)

If $\overrightarrow a = \widehat i + \widehat j + \widehat k$, $\overrightarrow b = 2\widehat i - \widehat j + 3\widehat k$ and $\overrightarrow c = \widehat i - 2\widehat j + \widehat k$, find a unit vector parallel to the vector $2\overrightarrow a - \overrightarrow b + 3\overrightarrow c $

Solution

$\overrightarrow a = \widehat i + \widehat j + \widehat k$, $\overrightarrow b = 2\widehat i - \widehat j + 3\widehat k$ and $\overrightarrow c = \widehat i - 2\widehat j + \widehat k$,
$2\overrightarrow a - \overrightarrow b + 3\overrightarrow c = 2\left( {\widehat i + \widehat j + \widehat k} \right) - \left( {2\widehat i - \widehat j + 3\widehat k} \right) + 3\left( {\widehat i - 2\widehat j + \widehat k} \right)$
$ = \cancel{2\widehat i} + 2\widehat j + 2\widehat k - \cancel{ 2\widehat i} + \widehat j - \cancel{3\widehat k} + 3\widehat i - 6\widehat j + \cancel{ 3\widehat k}$
$2\overrightarrow a - \overrightarrow b + 3\overrightarrow c = 3\widehat i - 3\widehat j + 2\widehat k$
$\left| {2\overrightarrow a - \overrightarrow b + 3\overrightarrow c } \right| = \sqrt {{3^2} + {{\left( { - 3} \right)}^2} + {{\left( 2 \right)}^2}} $
$\left| {2\overrightarrow a - \overrightarrow b + 3\overrightarrow c } \right| = \sqrt {9 + 9 + 4} $
$\left| {2\overrightarrow a - \overrightarrow b + 3\overrightarrow c } \right| = \sqrt {22}$
Unit vector parallel to $2\overrightarrow a - \overrightarrow b - 3\overrightarrow c = \frac{{3\widehat i - 3\widehat j + 2\widehat k}}{{\sqrt {22} }}$
$ = \frac{3}{{\sqrt {22} }}\widehat i - \frac{3}{{\sqrt {22} }}\widehat j + \frac{2}{{\sqrt {22} }}\widehat k$

Question (8)

Show that the points A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC

Solution

A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7)
$\overrightarrow {AB} = \left( {5,0, - 2} \right) - \left( {1, - 2. - 8} \right)$
$\overrightarrow {AB} = \left( {4,2,6} \right)$
$\overrightarrow {AB} = 4\widehat i + 2\widehat i + 6\widehat k$
$\overrightarrow {AC} = \left( {11,3,7} \right) - \left( {1, - 2, - 8} \right)$
$\overrightarrow {AC} = \left( {10,5,15} \right)$
$\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\4&2&6\\{10}&5&{15}\end{array}} \right|$
$\overrightarrow {AB} \times \overrightarrow {AC} = 0\widehat i - 0\widehat j + 0\widehat k = \overrightarrow 0 $
⇒ $\overrightarrow {AB} \, and \,\overrightarrow {AC} $ are parallel vectors
But A is initial points of both vector
$\overrightarrow {AB} \, and \,\overrightarrow {AC} $ are collinear vectors
∴ A, B, C are collinear points Let B divides $\overrightarrow {AC} $ in ratio λ : 1

$\left( {5,0, - 2} \right) = \left( {\frac{{11\lambda + 1}}{{\lambda + 1}},\frac{{3\lambda - 2}}{{\lambda + 1}},\frac{{7\lambda - 8}}{{\lambda + 1}}} \right)$
$ \Rightarrow \frac{{3\lambda - 2}}{{\lambda + 1}} = 0$
$ \Rightarrow 3\lambda - 2 = 0$
$\lambda = \frac{2}{3}$
∴ B divides AC in ratio 2:3

Question (9)

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\left( {2\overrightarrow a + \overrightarrow b } \right)$ and $\left( {\overrightarrow a - 3\overrightarrow b } \right)$ externally in the ratio 1:2. Also, show that P is the mid point of the line segment RQ

Solution

$P\left( {\overrightarrow p } \right) = 2\overrightarrow a + \overrightarrow b \quad \;Q\left( {\overrightarrow q } \right) = \overrightarrow a - 3\overrightarrow b $
R divide $\overrightarrow {PQ} $ externally in ratio 1:2
$R\left( {\overrightarrow r } \right) = \frac{{ - 1\overrightarrow q + 2\overrightarrow q }}{{ - 1 + 2}}$
$R\left( {\overrightarrow r } \right) = \frac{{ - 1\left( {\overrightarrow a - 2\overrightarrow b } \right) + 2\left( {2\overrightarrow a + \overrightarrow b } \right)}}{{ - 1}}$
$R\left( {\overrightarrow r } \right) = - \overrightarrow a + 3\overrightarrow b + 4\overrightarrow a + 2\overrightarrow b $
$R\left( {\overrightarrow r } \right) = 3\overrightarrow a + 5\overrightarrow b $

Question (10)

The two adjacent sides of a parallelogram are $2\widehat i - 4\widehat j + 5\widehat k$ and $\widehat i - 2\widehat j - 3\widehat k$. Find the unit vector parallel to its diagonal. Also, find its area.

Solution


$\overrightarrow {BC} = \widehat i - 2\widehat j - 3\widehat k \quad \overrightarrow {CD} = 2\widehat i - 4\widehat j + 5\widehat k$
$\overrightarrow {BD} = \widehat i - 2\widehat j - 3\widehat k + 2\widehat i - 4\widehat j + 5\widehat k$
$\overrightarrow {BD} = 3\widehat i - 6\widehat j + 2\widehat k$
unit vector $ = \frac{{3\widehat i - 6\widehat j + 2\widehat k}}{{\sqrt {9 + 36 + 4} }}$
unit vector $ = \frac{3}{7}\widehat i - \frac{6}{7}\widehat j + \frac{2}{7}\widehat k$
Area of parallelogram $ = \left| {\overrightarrow {BC} \times \overrightarrow {CD} } \right|$
$ = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\1&{ - 2}&{ - 3}\\2&{ - 4}&5\end{array}} \right|$
$ = - 22\widehat i - 11\widehat j + 0\widehat k$ $\left| {\overrightarrow {BC} \times \overrightarrow {CD} } \right| = \sqrt {{{\left( {22} \right)}^2} + {{\left( {11} \right)}^2}} $ $\left| {\overrightarrow {BC} \times \overrightarrow {CD} } \right| = \sqrt {484 + 121} $
$\left| {\overrightarrow {BC} \times \overrightarrow {CD} } \right| = \sqrt {605} = 11\sqrt 5 $
Area of parallelogram = $11\sqrt 5 $

Question (11)

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}$

Solution

Let $\overrightarrow a $ be vector which is equally inclined to axes
α, β γ are angle made by $\overrightarrow a $ with coordinate axes respectively and it is equally inclines
∴ α = β = γ
cos2α + cos2β + cos2γ = 1
cos2α +cos2α +cos2α =1
3cos2α =1
cosα = $\frac{1}{{\sqrt 3 }}$
Direction cosine of vector = (cosα, cosβ cosγ) = $\left( {\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}} \right)$

Question (12)

Let $\overrightarrow a = \widehat i + 4\widehat j + 2\widehat k$, $\overrightarrow b = 3\widehat i - 2\widehat j + 7\widehat k$ and $\overrightarrow c = 2\widehat i - \widehat j + 4\widehat k$. Find a vector $\overrightarrow d $ which is perpendicular to both $\overrightarrow a $ and $\overrightarrow b $, and $\overrightarrow {c\,} \cdot \,\overrightarrow d = 15$

Solution

$\overrightarrow a = \widehat i + 4\widehat j + 2\widehat k$, $\overrightarrow b = 3\widehat i - 2\widehat j + 7\widehat k$ and $\overrightarrow c = 2\widehat i - \widehat j + 4\widehat k$.
Let $\overrightarrow d $ is vector perpendicular to $\overrightarrow a $ and $\overrightarrow b $
$\overrightarrow d = \lambda \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat i}&{\widehat k}\\1&4&2\\3&{ - 2}&7\end{array}} \right|$
$\overrightarrow d = \lambda \left( {32\widehat i - \widehat j - 14\widehat k} \right)$
$\overrightarrow d = 32\lambda \widehat i - \lambda \widehat j - 14\lambda \widehat k$
Now $\overrightarrow c \cdot \overrightarrow d = 15$
$ \therefore \left( {2\widehat i - \widehat j + 4\widehat k} \right) \cdot \left( {32\lambda \widehat i - \lambda \widehat j - 14\lambda \widehat k} \right) = 15$
$64\lambda + \lambda - 56\lambda = 15$
$9\lambda = 15$
$\lambda = \frac{{15}}{9} = \frac{5}{3}$
$\overrightarrow d = 32\lambda \widehat i - \lambda \widehat j - 14\lambda \widehat k$
$\overrightarrow d = 32\left( {\frac{5}{3}} \right)\widehat i - \left( {\frac{5}{3}} \right)\widehat j - 14\left( {\frac{5}{3}} \right)\widehat k$
$\overrightarrow d = \frac{{160}}{5}\widehat i - \frac{5}{3}\widehat j - \frac{{70}}{3}\widehat k$

Question (13)

The scalar product of the vector $\widehat i + \widehat j + \widehat k$ with a unit vector along the sum of vectors $2\widehat i + 4\widehat j - 5\widehat k$ and $\lambda \widehat i + 2\widehat j + 3\widehat k$ is equal to one. Find the value of λ

Solution

$\overrightarrow a = \widehat i + \widehat j + \widehat k$,    $\overrightarrow b = 2\widehat i + 4\widehat j - 5\widehat k$,  and $\overrightarrow c = \lambda \widehat i + 2\widehat j + 3\widehat k$
$\overrightarrow b + \overrightarrow c = \left( {2 + \lambda } \right)\widehat i + 6\widehat i - 2\widehat k$
$\left| {\overrightarrow b + \overrightarrow c } \right| = \sqrt {{{\left( {2 + \lambda } \right)}^2} + {6^2} + {2^2}}$
$\left| {\overrightarrow b + \overrightarrow c } \right| = \sqrt {4 + 4\lambda + {\lambda ^2} + 36 + 4} $
$\left| {\overrightarrow b + \overrightarrow c } \right| = \sqrt {{\lambda ^2} + 4\lambda + 44} $
$\overrightarrow a \cdot \left( {\overrightarrow b + \overrightarrow c } \right) = 1$
$\left( {\widehat i + \widehat j + \widehat k} \right)\frac{{\left[ {\left( {2 + \lambda } \right)\widehat i + 6\widehat j - 2\widehat k} \right]}}{{\sqrt {{\lambda ^2} + 4\lambda + 44} }} = 1$
$\frac{{2 + \lambda + 6 - 2}}{{\sqrt {{\lambda ^2} + 4\lambda + 44} }} = 1$
$\lambda + 6 = \sqrt {{\lambda ^2} + 4\lambda + 44} $
Squaring on both sides
${\lambda ^2} + 12\lambda + 36 = {\lambda ^2} + 4\lambda + 44$
$8\lambda = 8$
\[\lambda = 1\]

Question (14)

If $\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $ are mutually perpendicular vectors of equal magnitude, show that the vector $\overrightarrow a + \,\overrightarrow b + \,\overrightarrow c $ is equally inclined to $\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $

Solution

$\overrightarrow a ,\,\overrightarrow b ,\,\overrightarrow c $ are mutually perpendicular vectors of equal magnitude
$ \therefore \overrightarrow a \cdot \overrightarrow b = \overrightarrow b \cdot \overrightarrow c = \overrightarrow c \cdot \overrightarrow a = 0$
$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| = k$ (say)
${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2} = \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) \cdot \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$
$ = \overrightarrow a \cdot \overrightarrow a + \overrightarrow a \cdot \overrightarrow b + \overrightarrow a \cdot \overrightarrow c + \overrightarrow b \cdot \overrightarrow a + \overrightarrow b \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a + \overrightarrow c \cdot \overrightarrow b + \overrightarrow c \cdot \overrightarrow c $
$ = {\left| {\overrightarrow a } \right|^2} + 0 + 0 + 0 + {\left| {\overrightarrow b } \right|^2} + 0 + 0 + 0 + {\left| {\overrightarrow c } \right|^2}$
$ = {k^2} + {k^2} + {k^2}$
$ \therefore \left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right| = \sqrt 3 k$ Let α be angle between $\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$ and ${\overrightarrow a }$ and ${\overrightarrow a }$ then
$\cos \alpha = \frac{{\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) \cdot \overrightarrow a }}{{\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|\left| {\overrightarrow a } \right|}}$
$\cos \alpha = \frac{{\overrightarrow a \cdot \overrightarrow a + \overrightarrow b \cdot \overrightarrow a + \overrightarrow c \cdot \overrightarrow a }}{{\sqrt 3 k \cdot k}}$
$\cos \alpha = \frac{{{{\left| {\overrightarrow a } \right|}^2} + 0 + 0}}{{\sqrt 3 {k^2}}}$
$\cos \alpha = \frac{{{k^2}}}{{\sqrt 3 {k^2}}}$
$\cos \alpha = \frac{1}{{\sqrt 3 }}$
$\alpha = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$
Similarly β and γbe angle mage by $\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$ with ${\overrightarrow b }$ and ${\overrightarrow c }$ respectively then we get
β = γ = cos-1 (1/√3)
$ \ therefore \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)$ is equally inclined with the vectors ${\overrightarrow a }$, ${\overrightarrow b }$, ${\overrightarrow c }$

Question (15)

Prove that $\left( {\overrightarrow a + \,\overrightarrow b } \right) \cdot \,\left( {\overrightarrow a + \,\overrightarrow b } \right) = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2}$ , if and only if $\overrightarrow a $, $\overrightarrow b $ are perpendicular, given $\overrightarrow a \ne \overrightarrow 0 ,\overrightarrow b \ne \overrightarrow 0 $

Solution

$\left( {\overrightarrow a + \overrightarrow b } \right) \cdot \left( {\overrightarrow a + \overrightarrow b } \right) = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2}$
$ \Rightarrow \overrightarrow a \cdot \overrightarrow a + \overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow a + \overrightarrow b \cdot \overrightarrow b = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2}$
$ \Rightarrow {\left| {\overrightarrow a } \right|^2} + 2\overrightarrow a \cdot \overrightarrow b + {\left| {\overrightarrow b } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2}$
$ \Rightarrow 2\overrightarrow a \cdot \overrightarrow b = 0$
$ \Rightarrow \overrightarrow a \cdot \overrightarrow b = 0$
$ \Rightarrow \overrightarrow a \bot \overrightarrow b $

Choose the correct answer in Exercise 16 to 19

Question (16)

If θ is the angle between two vectors $\overrightarrow a $ and $\overrightarrow b $, then $\overrightarrow a \cdot \overrightarrow b \ge 0$ only when
(A) $0 < \theta < \frac{\pi }{2}$
(B) $0 \le \theta \le \frac{\pi }{2}$
(C) $0 < \theta < \pi $
(D) $0 \le \theta \le \pi$

Solution

Let θ be angle between $\overrightarrow a $ and $\overrightarrow b $ then
$\cos \theta = \frac{{\overrightarrow a - \overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}$
$\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta = \overrightarrow a - \overrightarrow b $
$\overrightarrow a - \overrightarrow b \ge 0$
$\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta \ge 0$
$\cos \theta \ge 0$
θ lies in 1st quadrant
$ \therefore 0 \le \theta \le \frac{\pi }{2}$
∴ Option B is correct

Question (17)

Let $\overrightarrow a $ and $\overrightarrow a $ be two unit vectors and θ is the angle between them. Then $\overrightarrow a + \overrightarrow b $ is a unit vector if
(A) $\theta = \frac{\pi }{4}$
(B) $\theta = \frac{\pi }{3}$
(C) $\theta = \frac{\pi }{2}$
(D) $\theta = \frac{2\pi }{3}$

Solution

Given $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = 1$ and $\left| {\overrightarrow a + \overrightarrow b } \right| = 1$
${\left| {\overrightarrow a + \overrightarrow b } \right|^2} = \left( {\overrightarrow a + \overrightarrow b } \right) \cdot \left( {\overrightarrow a + \overrightarrow b } \right)$
${\left( 1 \right)^2} = \overrightarrow a \cdot \overrightarrow a + \overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow a + \overrightarrow b \cdot \overrightarrow b $
$1 = {\left| {\overrightarrow a } \right|^2} + 2\overrightarrow a \cdot \overrightarrow b + {\left| {\overrightarrow b } \right|^2}$
$1 = 1 + 2\overrightarrow a \cdot \overrightarrow b + 1$
$\overrightarrow a \cdot \overrightarrow b = \frac{{ - 1}}{2}$
$\left| {\overrightarrow a } \right| \cdot \left| {\overrightarrow b } \right|\cos \theta = \frac{{ - 1}}{2}$
$\cos \theta = \frac{{ - 1}}{2}$
$\theta = {\cos ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right)$
$\theta = \pi - {\cos ^{ - 1}}\left( {\frac{1}{2}} \right)$
$\theta = \pi - \frac{\pi }{3}$
$\theta = \frac{{2\pi }}{3}$
So option D is correct

Question (18)

The value of $\widehat i \cdot \left( {\widehat j \times \widehat k} \right) + \widehat j \cdot \left( {\widehat j \times \widehat k} \right) + \widehat k \cdot \left( {\widehat i \times \widehat j} \right)$ is
(A) 0     (B) -1
(C) 1     (D) 3

Solution

$\widehat i \cdot \left( {\widehat j \times \widehat k} \right) + \widehat j \cdot \left( {\widehat i \times \widehat k} \right) + \widehat k \cdot \left( {\widehat i \times \widehat j} \right)$
$ = \widehat i \cdot \widehat i + \widehat j \cdot \left( { - \widehat j} \right) + \widehat k \cdot \widehat k$
$ = {\left| {\widehat i} \right|^2} - {\left| {\widehat j} \right|^2} + {\left| {\widehat k} \right|^2}$
$ = {\left| {\widehat k} \right|^2 =1}$
∴ so option "C" is correct option

Question (19)

If θ is the angle between any two vectors $\overrightarrow a $ and $\overrightarrow b $, then $\left| {\overrightarrow a \cdot \overrightarrow b } \right| = \left| {\overrightarrow a \times \overrightarrow b } \right|$ when θ is equal to
(A) 0     (B) $\frac{\pi }{4}$
(C) $\frac{\pi }{2}$     (D) π

Solution

$\left| {\overrightarrow a \cdot \overrightarrow b } \right| = \left| {\overrightarrow a \times \overrightarrow b } \right|$
$ \Rightarrow \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta $
$ \Rightarrow \cos \theta = \sin \theta $
$ \Rightarrow \tan \theta = 1$
$ \Rightarrow \theta = {\tan ^{ - 1}}\left( 1 \right)$
$\Rightarrow \theta = \frac{\pi }{4}$
Option "B" is correct answer
Exercise 10.5 ⇐
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