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Write down a unit vector in the XY-plane, making an angle of ${30^\circ }$ with the positive direction of the x-axis.
Ans: Unit vector is $\vec r = \cos \theta \hat i + \sin \theta \hat j$, where $\theta $ is angle with positive ${\text{X}}$ axis. $\vec r = \cos {30^\circ }\hat i + \sin {30^\circ }\hat j = \dfrac{{\sqrt 3 }}{2}\hat i + \dfrac{1}{2}\hat j$
Find the scalar components and magnitude of the vector joining the points ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ and ${\text{Q}}\left( {{x_2},{y_2},{z_2}} \right)$
Ans: $\overrightarrow {{\text{PQ}}} = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k$
$|\overrightarrow {{\text{PQ}}} |\; = \;\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $
A girl walks 4 km towards the west, then she walks 3 km in a direction ${30^\circ }$ east of north and stops. Determine the girl's displacement from her initial point of departure.
Ans:
$\overrightarrow {{\text{OA}}} = - 4\hat i$
$\overrightarrow {{\text{AB}}} = \hat i|\overrightarrow {{\text{AB}}} |\cos {60^\circ } + \hat j|\overrightarrow {{\text{AB}}} |\sin {60^\circ }$
$ = \hat i3 \times \dfrac{1}{2} + \hat j3 \times \dfrac{{\sqrt 3 }}{2}$
$ = \dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$
$\overrightarrow {{\text{OB}}} = \overrightarrow {{\text{OA}}} + \overrightarrow {{\text{AB}}} $
$ = ( - 4\hat i) + \left( {\dfrac{3}{2}\hat i + \dfrac{{3\sqrt 3 }}{2}j} \right)$
$ = \left( { - 4 + \dfrac{3}{2}} \right)\hat i + \dfrac{{3\sqrt 3 }}{2}\hat j$
$ = \left( {\dfrac{{ - 8 + 3}}{9}} \right)\hat i + \dfrac{{3\sqrt 3 }}{9}\hat j$
$ = \dfrac{{ - 5}}{2}\vec i + \dfrac{{3\sqrt 3 }}{2}\hat j$
If $\overrightarrow a = \vec b + \vec c$, then is it true that $|\overrightarrow a | = |\vec b| + |\vec c|$ ? Justify your answer.
Ans: $\operatorname{In} \Delta {\text{ABC}},\overrightarrow {\;CB} = \vec a,\;\overrightarrow {CA} = b,\;\overrightarrow {AB} = \vec c$
$\vec a = \vec b + \vec c$, by triangle law of addition for vectors.
$|\vec a|\; < \;|\vec b| + |\vec c|$, by triangle inequality law of lengths.
Hence, it's not true that $|\vec a| = |\vec b| + |\vec c|$
Find the value of $x$ for which $x(\hat i + \hat j + \hat k)$ unit vector.
Ans: $|x(\hat i + \hat j + \hat k)| = 1$
$ \Rightarrow \sqrt {{x^2} + {x^2} + {x^2}} = 1$
$ \Rightarrow \sqrt {3{x^2}} = 1$
$ \Rightarrow \sqrt 3 x = 1$
$ \Rightarrow x = \pm \dfrac{1}{{\sqrt 3 }}$
Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec a = 2\hat i + 3\hat j - \hat k$ and $\vec b = \hat i - 2\hat j + \hat k$
Ans: $\vec c = \vec a + \vec b = (2 + 1)\hat i + (3 - 2)\hat j + ( - 1 + 1)\hat k = 3\hat i + \hat j$
$|\vec c| = \sqrt {{3^2} + {1^2}} = \sqrt {9 + 1} = \sqrt {10} $
$\therefore \hat c = \dfrac{{\vec c}}{{|\vec c|}} = \dfrac{{(3\hat i + \hat j)}}{{\sqrt {10} }}$
So, a vector of magnitude 5 and parallel to the resultant of $\vec a$ and $\vec b$ is $ \pm 5(\hat c) = \pm 5\left( {\dfrac{1}{{\sqrt {10} }}(3\hat i + \hat j)} \right) = \pm \dfrac{{3\sqrt {10} }}{2}\hat i \pm \dfrac{{\sqrt {10} }}{2}\hat j$
If $\vec a = \hat i + \hat j + \hat k\;,\;\vec b = 2\hat i - \hat j + 3\hat k$ and $\vec c = \hat i - 2\hat j + \hat k$, find a unit vector parallel to the vector $2\vec a - \vec b + 3\vec c$.
Ans: $2\vec a - \vec b + 3\vec c = 2(\hat i + \hat j + \hat k) - (2\hat i - \hat j + 3\hat k) + 3(\hat i - 2\hat j + \hat k)$
$ = 2\hat i + 2\hat j + 2\hat k - 2\hat i + j - 3\vec k + 3\hat i - 6\hat j + 3\hat k$
$ = 3\hat i - 3\hat j + 2\hat k$
$|2\vec a - \vec b + 3c| = \sqrt {{3^2} + {{( - 3)}^2} + {2^2}} = \sqrt {9 + 9 + 4} = \sqrt {22} $
Thus ,required unit vector is $\dfrac{{2\vec a - \vec b + 3\vec c}}{{|2\vec a - \vec b + 3\vec c|}} = \dfrac{{3\hat i - 3\hat j + 2\hat k}}{{\sqrt {22} }} = \dfrac{3}{{\sqrt {22} }}\hat i - \dfrac{3}{{\sqrt {22} }}\hat j + \dfrac{2}{{\sqrt {22} }}\hat k$
Show that the points ${\text{A}}(1, - 2, - 8),{\text{B}}(5,0, - 2)$ and ${\text{C}}(11,3,7)$ are collinear, and find the ratio in which B divides Ac.
Ans: $\overrightarrow {{\text{AB}}} = (5 - 1)\hat i + (0 + 2)\hat j + ( - 2 + 8)\hat k = 4\hat i + 2\hat j + 6\hat k$
$\overrightarrow {{\text{BC}}} = (11 - 5)\hat i + (3 - 0)\hat j + (7 + 2)\hat k = 6\hat i + 3\hat j + 9\hat k$
$\overrightarrow {AC} = (11 - 1)\hat i + (3 + 2)\hat j + (7 + 8)\hat k = 10\hat i + 5\hat j + 15\hat k$
$|\overrightarrow {{\text{AB}}} | = \sqrt {{4^2} + {2^2} + {6^2}} = \sqrt {16 + 4 + 36} = \sqrt {56} = 2\sqrt {14} $
$|\overrightarrow {{\text{BC}}} | = \sqrt {{6^2} + {3^2} + {9^2}} = \sqrt {36 + 9 + 81} = \sqrt {126} = 3\sqrt {14} $
$|\overrightarrow {AC} | = \sqrt {{{10}^2} + {5^2} + {{15}^2}} = \sqrt {100 + 25 + 225} = \sqrt {350} = 5\sqrt {14} $
$\therefore \;\;|\overrightarrow {{\text{AC}}} | = |\overrightarrow {{\text{AB}}} | + |\overrightarrow {{\text{BC}}} |$
So, the points are collinear.
Let B divide AC in ratio $\lambda :1$. $\overrightarrow {{\text{OB}}} = \dfrac{{\lambda \overrightarrow {{\text{OC}}} + \overrightarrow {{\text{OA}}} }}{{(\lambda + 1)}}$
$ \Rightarrow 5\hat i - 2\hat k = \dfrac{{\lambda (11\hat i + 3\hat j + 7\hat k) + (\hat i - 2\hat j - 8\hat k)}}{{\lambda + 1}}$
$ \Rightarrow (\lambda + 1)(5\hat i - 2\hat k) = 11\lambda \hat i + 3\lambda \hat j + 7\lambda \hat k + \hat i - 2\hat j - 8\hat k$
$ \Rightarrow 5(\lambda + 1)\hat i - 2(\lambda + 1)\hat k = (11\lambda + 1)\hat i + (3\lambda - 2)\hat j + (7\lambda - 8)\hat k$
$ \Rightarrow \lambda = \dfrac{2}{3}$
So, the required ratio is 2:3
Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2\vec a + \vec b)$ and $(\vec a - 3\vec b)$ externally in the ratio 1: 2. Also, show that $P$ is the midpoint of the line segment RQ.
Ans: $\overrightarrow {{\text{OP}}} = 2\vec a + \vec b,\overrightarrow {{\text{OQ}}} = \overrightarrow {\text{a}} - 3\overrightarrow {\text{b}} $
$\overrightarrow {OR} = \dfrac{{2(2\vec a + \vec b) - (\vec a - 3\vec b)}}{{2 - 1}} = \dfrac{{4\vec a + 2\vec b - \vec a - 3\vec b}}{1} = 3\vec a + 5\vec b$
So, the position vector of ${\text{R}}$ is $3\overrightarrow {\text{a}} + 5\overrightarrow {\text{b}} $
Position vector of midpoint of ${\text{RQ}} = \dfrac{{\overrightarrow {{\text{OQ}}} + \overrightarrow {{\text{OR}}} }}{2}$
$ = \dfrac{{(a\sqrt 6 ) + (3\vec a + 5\bar b)}}{2}$
$ = 2\vec a + \vec b$
$ = \overrightarrow {OP} $
Thus, ${\text{P}}$ is midpoint of line segment ${\text{RQ}}$
The two adjacent sides of a parallelogram are $2\hat i - 4\hat j + 5\hat k$ and $\hat i - 2\hat j - 3\hat k$. Find the unit vector parallel to its diagonal. Also, find its area.
Ans: Diagonal of a parallelogram is $\vec a + \vec b$
$\vec a + \vec b = (2 + 1)\hat i + ( - 4 - 2)\hat j + (5 - 3)\hat k = 3\hat i - 6\hat j + 2\hat k$
So, the unit vector parallel to diagonal is $\dfrac{{\vec a + \vec b}}{{|\vec a + \vec b|}} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {{3^2} + {{( - 6)}^2} + {2^2}} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{{\sqrt {9 + 36 + 4} }} = \dfrac{{3\hat i - 6\hat j + 2\hat k}}{7} = \dfrac{3}{7}\hat i - \dfrac{6}{7}\hat j + \dfrac{2}{7}\hat k$
$\vec{a}\times \vec{b} = \begin{vmatrix} \hat i& \hat j &\hat k \\ 2& -4 & 3\\ 1& -2 & -3 \end{vmatrix}$
$ = \hat i(12 + 10) - \hat j( - 6 - 5) + \bar k( - 4 + 4)$
$ = 22\hat i + 11\hat j$
$ = 11(2\hat i + \hat j)$
$\therefore |\vec a \times \vec b| = 11\sqrt {{2^2} + {1^2}} = 11\sqrt 5 $
So, the area of the parallelogram is $11\sqrt 5 $ sq units