12th NCERT Vector Algebra Miscellineous Questions 19

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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 1^st 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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