12th NCERT Vector Algebra Exercise 10.3 Questions 18
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Question (1)

Find the angle between two vectors $\overrightarrow a$ and $\overrightarrow b$ with magnitude $\sqrt 3$ and 2, respectively having
$\overrightarrow a \cdot \overrightarrow b = \sqrt 6$

Solution

$\left| {\overrightarrow a } \right| = \sqrt 3 ,\left| {\overrightarrow b } \right| = 2$ $\overrightarrow a \cdot \overrightarrow b = \sqrt 6$ Let θ be angle between $\overrightarrow a$ and $\overrightarrow b$ then $\cos \theta = \frac{{\overrightarrow a \cdot \overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}$ $\cos \theta = \frac{{\sqrt 6 }}{{\sqrt 3 \cdot 2}}$ $\cos \theta = \frac{{\sqrt 2 }}{2} = \frac{1}{{\sqrt 2 }}$ $\theta = \cos \left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4}$

Question (2)

Find the angle between the vectors $\widehat i - 2\widehat j + 3\widehat k$ and $3\widehat i - 2\widehat j + \widehat k$

Solution

$\overrightarrow a = \widehat i - 2\widehat j + 3\widehat k$
$\overrightarrow b = 3\widehat i - 2\widehat j + \widehat k$
Let θ be angle between $\overrightarrow a$ and $\overrightarrow b$
$\cos \theta = \frac{{\left( {\widehat i - 2\widehat j + 3\widehat k} \right) \cdot \left( {3\widehat i - 2\widehat j + \widehat k} \right)}}{{\sqrt {1 + 4 + 9} \cdot \sqrt {9 + 4 + 1} }}$ $\cos \theta = \frac{{3 + 4 + 3}}{{\sqrt {14} \cdot \sqrt {14} }}$ $\cos \theta = \frac{{\require{cancel} \cancel{10}^5}}{{\cancel{14}_7}}$ $\cos \theta = \frac{5}{7}$ $\theta = {\cos ^{ - 1}}\left( {\frac{5}{7}} \right)$

Question (3)

Find the projection of the vector $\widehat i - \widehat j$ on vector $\widehat i + \widehat j$

Solution

projection of the vector $\widehat i - \widehat j$ on vector $\widehat i + \widehat j = \frac{{\overrightarrow a \cdot \overrightarrow b }}{{\left| {\overrightarrow b } \right|}}$
$\overrightarrow a = \widehat i - \widehat j$ and $\overrightarrow b = \widehat i + \widehat j$
$projection = \frac{{\left( {\widehat i - \widehat j} \right) \cdot \left( {\widehat i + \widehat j} \right)}}{{\sqrt {1 + 1} }}$ $projection = \frac{{1 - 1}}{{\sqrt 2 }} = 0$

Question (4)

Find the projection of the vector $\widehat i + 3\widehat j + 7\widehat k$ on the vector $7\widehat i - \widehat j + 8\widehat k$

Solution

$\overrightarrow a = \hat i + 3\hat j + 7\hat k$ and $\overrightarrow b = 7\hat i - \hat j + 8\hat k$
Projection of $\overrightarrow a on \overrightarrow b = \frac{{\overrightarrow a \cdot \overrightarrow b }}{{\left| {\overrightarrow b } \right|}}$ $= \frac{{\left( {\widehat i + 3\widehat j + 7\widehat k} \right) \cdot \left( {7\widehat i - \widehat j + 8\widehat k} \right)}}{{\sqrt {49 + 1 + 64} }}$ $= \frac{{7 - 3 + 56}}{{\sqrt {114} }} = \frac{{60}}{{\sqrt {114} }}$

Question (5)

Show that each of the given three vectors is a unit vector: $\frac{1}{7}\left( {2\widehat i + 3\widehat j + 6\widehat k} \right)$ $\frac{1}{7}\left( {3\widehat i - 6\widehat j + 2\widehat k} \right)$ $\frac{1}{7}\left( {6\widehat i + 2\widehat j - 3\widehat k} \right)$ Also, show that they are mutually perpendicular to each other

Solution

$\overrightarrow a = \frac{1}{7}\left( {2\hat i + 3\hat j + 6\hat k} \right)$ $\left| {\overrightarrow a } \right| = \sqrt {\frac{{4 + 9 + 36}}{{49}} = 1}$ $\overrightarrow b = \frac{1}{7}\left( {3\hat i - 6\hat j + 2\hat k} \right)$ $\left| {\overrightarrow b } \right| = \sqrt {\frac{{9 + 36 + 4}}{{49}}} = 1$ $\overrightarrow c = \frac{1}{7}\left( {6\hat i + 2\hat j - 3\hat k} \right)$ $\left| {\overrightarrow c } \right| = \sqrt {\frac{{36 + 4 + 9}}{{49}}} = 1$ $\text {Now} \quad \overrightarrow a \cdot \overrightarrow b = \frac{1}{7}\left( {2\widehat i + 3\widehat j + 6\widehat k} \right) \cdot \frac{1}{7}\left( {3\widehat i - 6\widehat j + 2\widehat k} \right)$ $\overrightarrow a \cdot \overrightarrow b = \frac{1}{{49}}\left( {6 - 18 + 12} \right) = 0$ ⇒ $\overrightarrow a$ ⊥ $\overrightarrow b$
$\overrightarrow b \cdot \overrightarrow c = \frac{1}{7}\left( {3\widehat i - 6\widehat j + 2\widehat k} \right) \cdot \frac{1}{7}\left( {6\widehat i + 2\widehat j - 3\widehat k} \right)$ $\overrightarrow b \cdot \overrightarrow c = \frac{1}{{49}}\left( {18 - 12 - 6} \right) = 0$ ⇒ $\overrightarrow b$ ⊥ $\overrightarrow c$
$\overrightarrow a \cdot \overrightarrow c = \frac{1}{7}\left( {2\widehat i + 3\widehat j + 6\widehat k} \right) \cdot \frac{1}{7}\left( {6\widehat i + 2\widehat j - 3\widehat k} \right)$ $\overrightarrow a \cdot \overrightarrow c = \frac{1}{{49}}\left( {12 + 6 - 18} \right) = 0$ ⇒ $\overrightarrow a$ ⊥ $\overrightarrow c$
∴ vectors are mutually perpendicular

Question (6)

Find $\left| {\overrightarrow a } \right|$ and $\left| {\overrightarrow b } \right|$, if $\left( {\overrightarrow a + \overrightarrow b } \right) \cdot \left( {\overrightarrow a - \overrightarrow b } \right) = 8$ and $\left| {\overrightarrow a } \right| = 8\left| {\overrightarrow b } \right|$

Solution

$\left( {\overrightarrow a + \overrightarrow b } \right) \cdot \left( {\overrightarrow a - \overrightarrow b } \right) = 8$ $\overrightarrow a \cdot \overrightarrow a - \overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow a - \overrightarrow b \cdot \overrightarrow b = 8$ ${\left| {\overrightarrow a } \right|^2} - {\left| {\overrightarrow b } \right|^2} = 8$ ${\left( {8\left| {\overrightarrow b } \right|} \right)^2} - {\left| {\overrightarrow b } \right|^2} = 8$ $64{\left| {\overrightarrow b } \right|^2} - {\left| {\overrightarrow b } \right|^2} = 8$ $63{\left| {\overrightarrow b } \right|^2} = 8$ ${\left| {\overrightarrow b } \right|^2} = \frac{8}{{63}}$ $\left| {\overrightarrow b } \right| = \sqrt {\frac{8}{{63}}} = \frac{2}{3}\sqrt {\frac{2}{7}}$

Question (7)

Evaluate the product $\left( {3\overrightarrow a - 5\overrightarrow b } \right) \cdot \left( {2\overrightarrow a + 7\overrightarrow b } \right)$

Solution

$\left( {3\overrightarrow a - 5\overrightarrow b } \right) \cdot \left( {2\overrightarrow a + 7\overrightarrow b } \right)$ $= 6\overrightarrow a \cdot \overrightarrow a + 21\overrightarrow a \cdot \overrightarrow b - 10\overrightarrow b \cdot \overrightarrow a - 35\overrightarrow b \cdot \overrightarrow b$ $= 6{\left| {\overrightarrow a } \right|^2} + 11\overrightarrow a \cdot \overrightarrow b - 35{\left| {\overrightarrow b } \right|^2}$

Question (8)

Find the magnitude of two vectors $\overrightarrow a$ and $\overrightarrow b$ , having the same magnitude and such that the angle between them is 60° and their scalar product is $\frac{1}{2}$

Solution

$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|, \quad \theta = {60^o}, \quad \overrightarrow a \cdot \overrightarrow b = \frac{1}{2}$ $\cos \theta = \frac{{\overrightarrow a \cdot \overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}$ $\cos 60 = \frac{{\frac{1}{2}}}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}$ $\frac{1}{2} = \frac{1}{2}\frac{1}{{{{\left| {\overrightarrow a } \right|}^2}}}$ ${\left| {\overrightarrow a } \right|^2} = 1$ $\left| {\overrightarrow a } \right| = 1$ $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right| = 1$

Question (9)

Find $\left| {\overrightarrow x } \right|$, if for a unit vector $\overrightarrow a$. $\left( {\overrightarrow x - \overrightarrow a } \right) \cdot \left( {\overrightarrow x + \overrightarrow a } \right) = 12$

Solution

$\left| {\overrightarrow a } \right| = 1$ $\left( {\overrightarrow x - \overrightarrow a } \right) \cdot \left( {\overrightarrow x + \overrightarrow a } \right) = 12$ $\overrightarrow x \cdot \overrightarrow x + \overrightarrow x \cdot \overrightarrow a - \overrightarrow a \cdot \overrightarrow x - \overrightarrow a \cdot \overrightarrow a = 12$ ${\left| {\overrightarrow x } \right|^2} - {\left| {\overrightarrow a } \right|^2} = 12$ ${\left| {\overrightarrow x } \right|^2} - 1 = 12$ ${\left| {\overrightarrow x } \right|^2} = 13$ $\left| {\overrightarrow x } \right| = \sqrt {13}$

Question (10)

$\overrightarrow a = 2\widehat i + 2\widehat j + 3\widehat k$
$\overrightarrow b = - \widehat i + 2\widehat j + \widehat k$ and
$\overrightarrow c = 3\widehat i + \widehat j$
are such that $\overrightarrow a + \lambda \overrightarrow b$ is perpendicular to $\overrightarrow c$, then find value of λ

Solution

$\overrightarrow a = 2\widehat i + 2\widehat j + 3\widehat k$ $\overrightarrow b = - \widehat i + 2\widehat j + \widehat k$ $\text{As} \quad \overrightarrow a + \lambda \overrightarrow b \bot \overrightarrow c$ $\left( {\overrightarrow a + \lambda \overrightarrow b } \right) \cdot \overrightarrow c = 0$ $\overrightarrow a + \lambda \overrightarrow b = \left( {2\widehat i + 2\widehat j + 3\widehat k} \right) + \lambda \left( { - \widehat i + 2\widehat j + \widehat k} \right)$
$\left( {\overrightarrow a + \lambda \overrightarrow b } \right) \cdot \overrightarrow c = 0$ $\left[ {\left( {2 - \lambda } \right)\widehat i + \left( {2 + 2\lambda } \right)\widehat j + \left( {3 + \lambda } \right)\widehat k} \right] \cdot \left[ {3\widehat i + \widehat j} \right] = 0$ $3\left( {2 - \lambda } \right) + \left( {2 + 2\lambda } \right) = 0$ $6 - 3\lambda + 2 + 2\lambda = 0$ $8 - \lambda = 0$ $\lambda = 8$

Question (11)

Show that $\left| {\overrightarrow a } \right|\overrightarrow b + \left| {\overrightarrow b } \right|\overrightarrow a$ is perpendicular to $\left| {\overrightarrow a } \right|\overrightarrow b - \left| {\overrightarrow b } \right|\overrightarrow a$, for any two nonzero vector $\overrightarrow a$ and $\overrightarrow a$

Solution

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

Question (12)

If $\overrightarrow a \cdot \overrightarrow a = 0$ and $\overrightarrow a \cdot \overrightarrow b = 0$ , then what can be cocluded about the vector $\overrightarrow b$ ?

Solution

$\overrightarrow a \cdot \overrightarrow a = 0, \quad \overrightarrow a \cdot \overrightarrow b = 0$ ${\left| {\overrightarrow a } \right|^2} = 0, \quad 0 \cdot \overrightarrow b = 0$ $\text{since} \quad \overrightarrow a = \overrightarrow 0 ,\quad \overrightarrow b \text{can be any vector}$

Question (13)

If $\overrightarrow a$, $\overrightarrow b$, $\overrightarrow c$ are uniit vectors such that $\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow 0$, find the value of
$\overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a$

Solution

$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$ ${\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)^2} = 0$ $\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) \cdot \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) = 0$ $\overrightarrow a \cdot \overrightarrow a + \overrightarrow a \cdot \overrightarrow b + \overrightarrow a \cdot \overrightarrow c + \overrightarrow b \cdot \overrightarrow a + \overrightarrow b \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a + \overrightarrow c \cdot \overrightarrow b + \overrightarrow c \cdot \overrightarrow c = 0$ ${\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2\left( {\overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a } \right) = 0$ $1 + 1 + 1 + 2\left( {\overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a } \right) = 0$ $\overrightarrow a \cdot \overrightarrow b + \overrightarrow b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a = \frac{{ - 3}}{2}$

Question (14)

If either vector $\overrightarrow a = \overrightarrow 0$ or $\overrightarrow b = \overrightarrow 0$, then $\overrightarrow a \cdot \overrightarrow b = 0$. But the converse need not be true. Justify your answer with an example

Solution

$\overrightarrow a = 0,\overrightarrow b = 0$ $\text{then} \quad \overrightarrow a \cdot \overrightarrow b = \left| {\overrightarrow a } \right| \cdot \left| {\overrightarrow b } \right|\cos \theta = 0$ $text{If} \quad \overrightarrow a \cdot \overrightarrow b = 0$ $\Rightarrow \left| {\overrightarrow a } \right| = 0 \quad OR\left| {\overrightarrow b } \right| = 0 \quad OR\overrightarrow a \bot \overrightarrow b$ so converse statement is not true

Question (15)

If the vertices A, B, C of a triangle ABC are (1, 2, 3), (-1, 0, 0), (0, 1,2), respectively, then find ∠ABC. [ ∠ABC is the angle between the vector $\overrightarrow {BA}$ and $\overrightarrow {BC}$]

Solution

A(1, 2, 3), B(-1, 0, 0), C(0, 1, 2) $\cos \angle ABC = \frac{{\overrightarrow {BA} \cdot \overrightarrow {BC} }}{{\left| {\overrightarrow {BA} } \right| \cdot \left| {\overrightarrow {BC} } \right|}}$ $\overrightarrow {BA} = \left( {1,2,3} \right) - \left( { - 1,0,0} \right)$ $\overrightarrow {BA} = \left( {2,2,3} \right)$ $\overrightarrow {BC} = \left( {0,1,2} \right) - \left( { - 1,0,0} \right)$ $\overrightarrow {BC} = \left( {1,1,2} \right)$ $\cos \angle ABC = \frac{{\overrightarrow {BA} \cdot \overrightarrow {BC} }}{{\left| {\overrightarrow {BA} } \right| \cdot \left| {\overrightarrow {BC} } \right|}}$ $\cos \angle ABC = \frac{{\left( {2,2,3} \right) \cdot \left( {1,1,2} \right)}}{{\sqrt {4 + 4 + 9} \sqrt {1 + 1 + 4} }}$ $\cos \angle ABC = \frac{{2 + 2 + 6}}{{\sqrt {17} \sqrt 6 }}$ $\cos \angle ABC = \frac{{10}}{{\sqrt {102} }}$ $\angle ABC = {\cos ^{ - 1}}\left( {\frac{{10}}{{\sqrt {102} }}} \right)$

Question (16)

Show that the point A(1, 2, 7), B(2, 6, 3) and C (3, 10, -1) are collinear

Solution

A(1, 2, 7), B(2, 6, 3) and C (3, 10, -1) $\overrightarrow {AB} = \left( {2,6,3} \right) - \left( {1,2,7} \right) = \left( {1,4, - 4} \right)$ $\overrightarrow {BC} = \left( {3,10, - 1} \right) - \left( {2,6,3} \right) = \left( {1,4, - 4} \right)$ $\overrightarrow {AB} = \overrightarrow {BC}$ $\Rightarrow \lambda = 1$ so vectors are collinear

Question (17)

Show that the vectors $2\widehat i - \widehat j + \widehat k$, $\widehat i - 3\widehat j - 5\widehat k$ and $3\widehat i - 4\widehat j - 4\widehat k$ form the verices of a right angled triangle.

Solution

$\overrightarrow a = 2\hat i - \hat j + \hat k$ $\overrightarrow b = \hat i - 3\hat j - 5\hat k$ $\overrightarrow c = 2\hat i - 4\hat j - 4\hat k$ $\overrightarrow a \cdot \overrightarrow b = \left( {2\widehat i - \widehat j + \widehat k} \right) \cdot \left( {\widehat i - 3\widehat i - 5\widehat k} \right)$ $\overrightarrow a \cdot \overrightarrow b = 2 + 3 + 5 = 0$ $\Rightarrow \overrightarrow a \bot \overrightarrow b$ So they are are side of right angle traingle

Question (18)

If $\overrightarrow a$ is a nonzero vector of magnitude 'a' and λ a nonzero scalar, then λ$\overrightarrow a$ is unit vector if
(A) λ = 1   (B) λ = -1
(C) a = |λ|   (D) $a = \frac{1}{{\left| \lambda \right|}}$

Solution

$\left| {\overrightarrow a } \right| = \lambda \overrightarrow a$ $\lambda = 1$ So A is a correct option