Question 13 Marks
Find the coordinates of the tip of the position vector which is equivalent to $\overrightarrow{\text{AB}}$, where the coordinates of A and B are (-1, 3) and (-2, 1) respectively.
AnswerLet O be the origin. Let P(x, y) be the required point. Then $\vec{\text{P}}$ is the tip of the position vector $\overrightarrow{\text{OP}}$ of the point P. We have,$\overrightarrow{\text{OP}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}$
And, $\overrightarrow{\text{AB}}$ = Position vector of B - Position vector of A $=\big(-2\hat{\text{i}}+\hat{\text{j}}\big)-\big(-\hat{\text{i}}+3\hat{\text{j}}\big)$ $=-2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{i}}-3\hat{\text{j}}$ $=-\hat{\text{i}}-2\hat{\text{j}}$ Given that $\overrightarrow{\text{OP}}=\overrightarrow{\text{AB}}$ So, $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}=-\hat{\text{i}}-2\hat{\text{j}}\Leftrightarrow\text{x}=-1,\ \text{y}=-2$ Hence, coordinated of the required point is (-1, -2)
View full question & answer→Question 23 Marks
Prove that the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector.
AnswerGiven: A regular octagon of eight sides with center O.
To show: $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}+\overrightarrow{\text{OE}}+\overrightarrow{\text{OF}}+\overrightarrow{\text{OG}}+\overrightarrow{\text{OH}}=\vec0$.
Proof: We know center of the regular octagon bisects all the diagonals passing through it.
$\overrightarrow{\text{OA}}=-\overrightarrow{\text{OE}},\ \overrightarrow{\text{OB}}=-\overrightarrow{\text{OF}},\ \overrightarrow{\text{OD}}=-\overrightarrow{\text{OH}}$ and $\overrightarrow{\text{OC}}=-\overrightarrow{\text{OG}}.$
$\Rightarrow\overrightarrow{\text{OA}}+\overrightarrow{\text{OE}}=\vec0,\ \overrightarrow{\text{OB}}+\overrightarrow{\text{OF}}=\vec0,\ \overrightarrow{\text{OD}}+\overrightarrow{\text{OH}}=\vec0$ and $\overrightarrow{\text{OC}}+\overrightarrow{\text{OG}}=\vec0.\ \dots(\text{i})$
Now,
$\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}+\overrightarrow{\text{OE}}+\overrightarrow{\text{OF}}+\overrightarrow{\text{OG}}+\overrightarrow{\text{OH}}$
$=\Big(\overrightarrow{\text{OA}}+\overrightarrow{\text{OE}}\Big)+\Big(\overrightarrow{\text{OB}}+\overrightarrow{\text{OF}}\Big)+\Big(\overrightarrow{\text{OC}}+\overrightarrow{\text{OG}}\Big)+\Big(\overrightarrow{\text{OD}}+\overrightarrow{\text{OH}}\Big)$
$=\vec0+\vec0+\vec0+\vec0$
$=\vec0$
Hence proved.
View full question & answer→Question 33 Marks
Find the position vector of a point R which divides the line joining the two points P and Q with position vectors $\overrightarrow{\text{OP}}=2\vec{\text{a}}+\vec{\text{b}}\text{ and }\overrightarrow{\text{OQ}}=\vec{\text{a}}-2\vec{\text{b}}$, respectively in the ratio 1 : 2 internally and externally.
AnswerIt is given that P and Q are two points with position vectors $\overrightarrow{\text{OP}}=2\vec{\text{a}}+\vec{\text{b}}\text{ and }\overrightarrow{\text{OQ}}=\vec{\text{a}}-2\vec{\text{b}}$, respectively.
When R divides PQ internally in the ratio 1 : 2, then
Position vector of $\text{R}=\frac{1\times\big(\vec{\text{a}}-2\vec{\text{b}}\big)+2\times\big(2\vec{\text{a}}+\vec{\text{b}}\big)}{1+2}=\frac{5\vec{\text{a}}}{3}$
When R divides PQ externally in the ratio 1 : 2, then
Position vector of $\text{R}=\frac{1\times\big(\vec{\text{a}}-2\vec{\text{b}}\big)-2\times\big(2\vec{\text{a}}+\vec{\text{b}}\big)}{1-2}=\frac{-3\vec{\text{a}}-4\vec{\text{b}}}{-1}=3\vec{\text{a}}+4\vec{\text{b}}$
View full question & answer→Question 43 Marks
Prove that the given vectors are coplanar:
$\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ 2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ and $-\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
View full question & answer→Question 53 Marks
ABCDE is a pentagon, prove that,$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}=3\ \overrightarrow{\text{AC}}$
AnswerIt is given that ABCDE is a pentagon, So
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}$
$=\Big(\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}\Big)+\overrightarrow{\text{AE}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\Big(\overrightarrow{\text{AE}}+\overrightarrow{\text{ED}}\Big)+\overrightarrow{\text{AC}}$ $\Big[$Using triangle law in $\triangle\text{ABC},\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\Big]$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\Big(\overrightarrow{\text{AD}}\Big)+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}-\overrightarrow{\text{DA}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{AD}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{AD}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{AC}}$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{AC}}+\overrightarrow{\text{AC}}$
$=3\ \overrightarrow{\text{AC}}$
So,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AE}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{DC}}+\overrightarrow{\text{ED}}+\overrightarrow{\text{AC}}=3\ \overrightarrow{\text{AC}}$
View full question & answer→Question 63 Marks
Find the vector from the origin O to the centroid of the triangle whose vertices are (1, -1, 2), (2, 1, 3) and (-1, 2, -1).
AnswerGiven the vertices of the triangle (1, -1, 2), (2, 1, 3) and (-1, 2, -1). Then, Position vectors are: $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ $\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ The centroid of a triangle is given by $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{3}$ So, $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{3}=\frac{\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}+2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}}3{}$ $=\frac{2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}}3=\frac{2}3\hat{\text{i}}+\frac{2}3\hat{\text{j}}+\frac{4}3\hat{\text{k}}$
View full question & answer→Question 73 Marks
Prove that the given vectors are coplanar:
$2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\ \hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}$ and $3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}$
AnswerGiven the vectors $\text{P}\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big),\ \text{Q}\big(\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}\big)$ and $\text{R}\big(3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}\big)$
We know the three vectors are coplanar if oneof them is expressible as a linear combination of the other two. Let,
$2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}=\text{x}\big(\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}\big)+\text{y}\big(3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}\big)$
$=\hat{\text{i}}(\text{x}+3\text{y})+\hat{\text{j}}(-3\text{x}-4\text{y})+\hat{\text{k}}(-5\text{x}-4\text{y})$
$\Rightarrow\text{x}+3\text{y}=2,\ -3\text{x}-4\text{y}=-1,\ -5\text{x}-4\text{y}=1$ [Equating the coefficients of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}$ respectively]
Solving first two of these equation, we get x = -1, y = 1. Clearly these two values satisfy the third equation.
Hence, the given vectors are coplanar.
View full question & answer→Question 83 Marks
ABCD is a parallelogram. If the coordinates of A, B, C are (-2, -1), (3, 0) and (1, -2) respectively, find the coordinates of D.
AnswerLet the coordinates of D is (x, y).
Since, ABCD is a parallelogram.
$\therefore$ AB = DC
We have,
$\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}$
$\Rightarrow3\hat{\text{i}}-\big(-2\hat{\text{i}}-\hat{\text{j}}\big)=\big(\hat{\text{i}}-2\hat{\text{j}}\big)-\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}\big)$
$\Rightarrow5\hat{\text{i}}+\hat{\text{j}}=\hat{\text{i}}(1-\text{x})+\hat{\text{j}}(-2-\text{y})$
$\Rightarrow1-\text{x}=5\text{ and }1=-2-\text{y}$
$\Rightarrow\text{x}=-4\text{ and }\text{y}=-3$
Hence, the coordinates of D is (-4, -3)
View full question & answer→Question 93 Marks
Show that the four points having position vectors $6\hat{\text{i}}-7\hat{\text{j}},\ 16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},\ 3\hat{\text{j}}-6\hat{\text{k}},\ 2\hat{\text{i}}-5\hat{\text{j}}+10\hat{\text{k}}$ are coplanar.
AnswerLet the given four points be P, Q, R and S respectively. Three points are coplanar if the vectors $\overrightarrow{\text{PQ}},\ \overrightarrow{\text{PR}}\text{ and }\overrightarrow{\text{PS}}$ are coplanar. These vectors are coplanar if one of them can be expressed as a linear combination of the other two. So, let$\overrightarrow{\text{PQ}}=\text{x}\overrightarrow{\text{PR}}+\text{y}\overrightarrow{\text{PS}}$
$10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}\\=\text{x}\big(-6\hat{\text{i}}+10\hat{\text{j}}-6\hat{\text{k}}\big)+\text{y}\big(-4\hat{\text{i}}+2\hat{\text{j}}+10\hat{\text{k}}\big)$
$\Rightarrow10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}\\=\hat{\text{i}}\big(-6{\text{x}}-4\text{y}\big)+\hat{\text{j}}\big(10{\text{x}}+2{\text{y}}\big)+\hat{\text{k}}\big(-6\text{x}+10\text{y}\big)$
$\Rightarrow-6\text{x}-4\text{y}=10,\ 10\text{x}+2\text{y}=-12$ and $-6\text{x}+10\text{y}=-4$ [Equating coefficients of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}$ on both sides]
Solving the first of these three equations, we get x = -1 and y = -1. These values also satisfy the thied equation. Hence, the given four points are coplanar.
View full question & answer→Question 103 Marks
Using vectors show that the points A(-2, 3, 5), B(7, 0, -1), C(-3. -2, -5) and D(3, 4, 7) are such that AB and CD intersect at the point P(1, 2, 3).
AnswerWe have,
$\overrightarrow{\text{AP}}=$ Position vector of P - Position vector of A
$\Rightarrow\overrightarrow{\text{AP}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)-\big(-2\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}\big)$
$=3\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$
$\overrightarrow{\text{PB}}=$ Position vector of B - Position vector of P
$\Rightarrow\overrightarrow{\text{PB}}=\big(7\hat{\text{i}}-0\hat{\text{j}}-\hat{\text{k}}\big)-\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=6\hat{\text{i}}-2\hat{\text{j}}-4\hat{\text{k}}$
Since $\overrightarrow{\text{PB}}=2\overrightarrow{\text{AP}}$. So, vectors $\overrightarrow{\text{PB}}$ and $\overrightarrow{\text{AP}}$ are collinear. But P is a point common to $\overrightarrow{\text{PB}}$ and $\overrightarrow{\text{AP}}$.
Hence P, A, B are collinear points.
Now, $\overrightarrow{\text{CP}}=\big(-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}\big)-\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=\big(-4\hat{\text{i}}-4\hat{\text{j}}-8\hat{\text{k}}\big)$
$\overrightarrow{\text{PD}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)-\big(3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}\big)$
$=\big(-2\hat{\text{i}}-2\hat{\text{j}}-4\hat{\text{k}}\big)$
Thus, $\overrightarrow{\text{CP}}=2\overrightarrow{\text{PD}}$
So the vectors $\overrightarrow{\text{CP}}$ and $\overrightarrow{\text{PD}}$ are collinear. But P is a common point to $\overrightarrow{\text{CP}}$ and $\overrightarrow{\text{PD}}$
Hence, C, P, D are collinear points.
Thus A, B, C, D and P are points such that A, P, B and C, P, D are two sets of collinear points.
Hence, AB and CD intersect at point P.
View full question & answer→Question 113 Marks
Write the position vector of the point which divdes the join of the points with position vectors $3\vec{\text{a}} - 2\vec{\text{b}} $ and $2\vec{\text{a}} + 3\vec{\text{b}} $ in the ratio 2 : 1.
AnswerSuppose R be the point which divdes the line joining the points with position vectors $3\vec{\text{a}} - 2\vec{\text{b}} $ and $2\vec{\text{a}} + 3\vec{\text{b}} $ in the ratio 2 : 1
And $\overrightarrow{\text{OA}} = 3\vec{\text{a}} - 2\vec{\text{b}} $ and $\overrightarrow{\text{OB}} = 2\vec{\text{a}} + 3\vec{\text{b}} $
Here, m : n = 2 : 1
Therefore, position vector $\overrightarrow{\text{OR}}$ is as follows:
$\overrightarrow{\text{OR}} = \frac{\text{m} \overrightarrow{\text{OB}} + \text{n} \overrightarrow{\text{OA}}}{\text{m + n}}$
$ = \frac{2\big(2\vec{\text{a}} + 3 \vec{\text{b}}\big) + 1 \big(3\vec{\text{a}} + 2 \vec{\text{b}}\big)}{2 + 1}$
$= \frac{7\vec{\text{a}} + 4\vec{\text{b}}}{3}$
$= \frac{7}{3} \vec{\text{a}} + \frac{4}{3} \vec{\text{b}}$
View full question & answer→Question 123 Marks
If $\overrightarrow{\text{PQ}}=3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ and the coordinates of P are (1, -1, 2), find the coordinates of Q.
AnswerHere, $\overrightarrow{\text{PQ}}=3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Position vector of $\text{P}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\overrightarrow{\text{PQ}}=$ Position vector of Q - Position vector of P
$3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}=$ Position vector of Q $-\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$
Position vector of $\text{Q}=\big(\hat{3\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)+\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$
$=4\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Coordinates of Q = (4, 1, 1)
View full question & answer→Question 133 Marks
A vector $\vec{\text{r}}$ is inclined at equal angles to the three axes. If the magnitude of $\vec{\text{r}}$ is $2\sqrt3$, find $\vec{\text{r}}$.
AnswerLet l, m, n be the direction cosines of $\vec{\text{r}}$.
Now, $\vec{\text{r}}$ is inclined at equal angles to the three axes.
$\therefore\ \text{l}=\text{m}=\text{n}$ $[\alpha=\beta=\gamma\Rightarrow\cos\alpha=\cos\beta=\cos\gamma]$
So, $\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ 3\text{l}^2=1$
$\Rightarrow\ \text{l}=\pm\frac{1}{\sqrt3}$
$\Rightarrow\ \text{l = m = n}=\pm\frac{1}{\sqrt3}$
We know that,
$\vec{\text{r}}=|\vec{\text{r}}|(\text{l}\hat{\text{i}}+\text{m}\hat{j}+\text{n}\hat{\text{k}})$
$\Rightarrow\ \vec{\text{r}}=2\sqrt3\Big(\pm\frac{1}{\sqrt3}\hat{\text{i}}\pm\frac{1}{\sqrt3}\hat{\text{j}}\pm\frac{1}{\sqrt3}\hat{\text{k}}\Big)$
$\Rightarrow\ \vec{\text{r}}=2\big(\pm\hat{\text{i}}\pm\hat{\text{j}}\pm\hat{\text{k}}\big)$
View full question & answer→Question 143 Marks
ABCD is a parallelogram and P is the point of intersection of its diagonals. If O is the origin of reference, show that $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}=4\ \overrightarrow{\text{OP}}$.
Answer
Given a parallelogram ABCD and P is the point of intersection of its diagonals. We know the diagonals of a parallelogram, bisect eacg other. Therefore,
$\frac{\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}}2=\overrightarrow{\text{OP}}$
$\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}=2\ \overrightarrow{\text{OP}}\ \dots(1)$
and $\frac{\overrightarrow{\text{OB}}+\overrightarrow{\text{OD}}}2=\overrightarrow{\text{OP}}$
$\overrightarrow{\text{OB}}+\overrightarrow{\text{OD}}=2\ \overrightarrow{\text{OP}}\ \dots(2)$
Adding (1) and (2), We get,
$\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}=4\ \overrightarrow{\text{OP}}$ View full question & answer→Question 153 Marks
In a triangle OAC, if B is the mid-point of side AC and $\overrightarrow{\text{OA}} = \vec{\text{a}}, \overrightarrow{\text{OB}} = \vec{\text{b}},$ then was is $\overrightarrow{\text{OC}}$?
AnswerIn $\triangle\text{OAC}, \overrightarrow{\text{OA}} = \vec{\text{a}} $ and $ \overrightarrow{\text{OB}} = \vec{\text{b}}.$
It is given that B is the mid-point of AC.
$\therefore$ Position vector of B $= \frac{\text{Positionn vector of A + Positionn vector of C}}{2}$
$\Rightarrow \overrightarrow{\text{OB}} = \frac{\overrightarrow{\text{OA}} + \overrightarrow{\text{OC}}}{2}$
$\Rightarrow \vec{\text{b}} = \frac{\vec{\text{a}}+\overrightarrow{\text{OC}}}{2}$
$\Rightarrow \vec{\text{a}} + \overrightarrow{\text{OC}} = 2\vec{\text{b}}$
$\Rightarrow \overrightarrow{\text{OC}} = 2\vec{\text{b}} - \vec{\text{a}}$ View full question & answer→Question 163 Marks
If the position vector $\vec{\text{a}}$ of a point (12, n) is such that $\big|\vec{\text{a}}\big|=13$, find the value (s) of n.
AnswerGiven a position vector $\vec{\text{a}}$ of a point (12, n) such that,
$\vec{\text{a}}=12\hat{\text{i}}+\text{n}\hat{\text{j}}$
Then,
$\big|\vec{\text{a}}\big|=\sqrt{12^2+\text{n}^2}$
Also, $\big|\vec{\text{a}}\big|=13$ (given)
Thus, we get,
$\sqrt{12^2+\text{n}^2}=13$
$\Rightarrow12^2+\text{n}^2=169$
$\Rightarrow\text{n}^2=169-144$
$\Rightarrow\text{n}^2=25$
$\Rightarrow\text{n}=\pm5$
View full question & answer→Question 173 Marks
If D, E, F are the mid-points of side BC, CA and AB respectively of a triangle ABC, write the value of $\overrightarrow{\text{AD}}+\overrightarrow{\text{BE}}+\overrightarrow{\text{CF}}$.
AnswerGiven: D, E, F are the mid-points of the sides BC, CA, AB respectively. Then, the position vectors of the mid-points D, E, F are given by $\frac{\vec{\text{b}}+\vec{\text{c}}}2,\ \frac{\vec{\text{c}}+\vec{\text{a}}}2,\ \frac{\vec{\text{a}}+\vec{\text{b}}}2$Now, $\overrightarrow{\text{AD}}+\overrightarrow{\text{BE}}+\overrightarrow{\text{CF}}$
$=\Big(\frac{\vec{\text{b}}+\vec{\text{c}}}2\Big)-\vec{\text{a}}+\Big(\frac{\vec{\text{c}}+\vec{\text{a}}}2\Big)-\vec{\text{b}}+\Big(\frac{\vec{\text{a}}+\vec{\text{b}}}2\Big)-\vec{\text{c}}$
$=2\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}2\Big)-\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)-\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\vec0$
View full question & answer→Question 183 Marks
Find a vector $\vec{\text{r}}$ of magnitude $3\sqrt{2}$ units which makes an angle of $\frac{\pi}{4}$ and $\frac{\pi}{2}$ with and z-axes respectively.
AnswerSuppose vector $\vec{\text{r}}$ makes an angle with the $\alpha$ x-axis.Let l, m, n be the direction cosines of $\vec{\text{r}}$. Then,
$\text{l}=\cos\alpha,\text{m}=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}},\text{n}=\cos\frac{\pi}{2}=0$
Now,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\cos^2\alpha+\frac{1}{2}+0=1$
$\Rightarrow\cos^2\alpha=1-\frac{1}{2}=\frac{1}{2}$
$\Rightarrow\cos^2\alpha=\pm\frac{1}{\sqrt{2}}$
We know that,
$\vec{\text{r}}=|\vec{\text{r}}|\big(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}}\big)$
$\therefore\vec{\text{r}}=3\sqrt{2}\Big(\pm\frac{1}{\sqrt{2}}\hat{\text{i}}+\frac{1}{\sqrt{2}}\hat{\text{i}}+0\hat{\text{k}}\Big)\ (|\vec{\text{a}}|=3\sqrt{2})$
$\Rightarrow\vec{\text{r}}=\pm3\hat{\text{i}}+3\hat{\text{j}}$
View full question & answer→Question 193 Marks
Find the unit vector in the direction of vector $\overrightarrow{\text{PQ}}$, where P and Q are the points (1, 2, 3) and (4, 5, 6).
AnswerLet $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are the position vectors of the point P(1, 2, 3) and Q(4, 5, 6)
Then,
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}$
So,
$\overrightarrow{\text{PQ}}=\vec{\text{b}}-\vec{\text{a}}$
$=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$=3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}$
Now, $\Big|\overrightarrow{\text{PQ}}\Big|=\sqrt{3^2+3^2+3^2}$
$=\sqrt{9+9+9}$
$=3\sqrt3$
Therefore, Unit vector parallel to $\overrightarrow{\text{PQ}}=\frac{\overrightarrow{\text{PQ}}}{|\text{PQ}|}=\frac{1}{3\sqrt3}\big(3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}\big)$
$=\frac{1}{\sqrt3}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→Question 203 Marks
A vector $\vec{\text{r}}$ is inclined to x-axis at 45º and y-axis at 60º. If $|\vec{\text{r}}|=8$ units, find $\vec{\text{r}}$.
AnswerHere, $\alpha=45^{\circ},\ \beta=60^{\circ},\ \gamma=\theta$ (say)
$\text{l}=\cos\alpha$
$=\cos45^{\circ}$
$\text{l}=\frac{1}{\sqrt2}$
$\text{m}=\cos\beta$
$=\cos60^{\circ}$
$\text{m}=\frac{1}2$
$\text{n}=\cos\theta$
Put l, m and n in,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Big(\frac{1}{\sqrt2}\Big)^2+\Big(\frac{1}2\Big)^2+\cos^2\theta=1$
$\frac{2+1}4+\cos^2\theta=1$
$\frac{3}4+\cos^2\theta=1$
$\cos^2\theta=\frac{1}1-\frac{3}4$
$=\frac{4-3}4$
View full question & answer→Question 213 Marks
Find the position vector of the mid-point of the vector joining the points $\text{P}\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)$ and $\text{Q}\big(4\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\big)$.
AnswerGiven: $\text{P}\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)$ and $\text{Q}\big(4\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\big)$The position vector of the mid-point of the vector
joining these points $=\frac{\text{Position vector of P}+\text{Position vector of Q}}{2}$
$=\frac{\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+\big(4\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\big)}{2}$
$=\frac{6\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{2}$
$=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
View full question & answer→Question 223 Marks
Show that the points whose position vectors are as given below are collinear:
$3\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}},\ \hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ and $-\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}$
AnswerLet the points be A, B and C with position vectors $3\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}},\ \hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ and $-\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}$ respectively. Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}-3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$
$=-2\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=-\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
$=-2\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, A, B, and C are collinear.
View full question & answer→Question 233 Marks
Find a vector of magnitude of 5 units parallel to the resultant of the vectors $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$.
AnswerGiven that, $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$Thus, Find a vector of mangnitude of 5 units parallel to the resultant of the vectors
$\vec{\text{a}}+\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}+\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{j}}$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{9+1}=\sqrt{10}$
Thus, the unit vector along the resultant vector $\vec{\text{a}}+\vec{\text{b}}$ is $\frac{3\hat{\text{i}}+\hat{\text{j}}}{\sqrt{10}}$
The vector of magnitude of 5 units parallel to the resultant vector$=\frac{3\hat{\text{i}}+\hat{\text{j}}}{\sqrt{10}}\times5=\sqrt{\frac{5}{2}}\big(3\hat{\text{i}}+\hat{\text{j}}\big)$
View full question & answer→Question 243 Marks
If the vertices A, B, C of a triangle ABC are the points with position vectors $\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}},\ \text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}},\ \text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}$ respectively, what are the vectors determined by its sides? Find the length of these vectors.
AnswerGiven the vertices of a triangle A, B and C with position vectors $\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}},\ \text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ and $\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}$ respectively. Then,
$\overrightarrow{\text{AB}}=(\text{b}_1-\text{a}_1)\hat{\text{i}}+(\text{b}_2-\text{a}_2)\hat{\text{j}}+(\text{b}_3-\text{a}_3)\hat{\text{k}}$
$\overrightarrow{\text{BC}}=(\text{c}_1-\text{b}_1)\hat{\text{i}}+(\text{c}_2-\text{b}_2)\hat{\text{j}}+(\text{c}_3-\text{b}_3)\hat{\text{k}}$
$\overrightarrow{\text{CA}}=(\text{a}_1-\text{c}_1)\hat{\text{i}}+(\text{a}_2-\text{c}_2)\hat{\text{j}}+(\text{a}_3-\text{c}_3)\hat{\text{k}}$
Therefore, the length of these vectors are:
$\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(\text{b}_1-\text{a}_1)^2+(\text{b}_2-\text{a}_2)^2+(\text{b}_3-\text{a}_3)^2}$
$\Big|\overrightarrow{\text{BC}}\Big|=\sqrt{(\text{c}_1-\text{b}_1)^2+(\text{c}_2-\text{b}_2)^2+(\text{c}_3-\text{b}_3)^2}$
$\Big|\overrightarrow{\text{CA}}\Big|=\sqrt{(\text{a}_1-\text{c}_1)^2+(\text{a}_2-\text{c}_2)^2+(\text{a}_3-\text{c}_3)^2}$
View full question & answer→Question 253 Marks
Give a condition that three vectors $\vec{\text{a}},\ \vec{\text{b}}\text{ and }\vec{\text{c}}$ from the three sides of a triangle. what are the other possibilities?
AnswerLet ABC be a triangle such that $\overrightarrow{\text{BC}}=\vec{\text{a}},\ \overrightarrow{\text{AB}}=\vec{\text{c}}\text{ and }\overrightarrow{\text{CA}}=\vec{\text{b}}$. Then,
$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}+\overrightarrow{\text{AB}}$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\overrightarrow{\text{BA}}+\overrightarrow{\text{AB}}$ $\big[\because\ \overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\overrightarrow{\text{BA}}\big]$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\overrightarrow{\text{BB}}$ [Using triangle law]
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec0$ [By defination of null vector]
Other possibilities are,
- $\vec{\text{c}}+\vec{\text{a}}=\vec{\text{b}}$
- $\vec{\text{a}}+\vec{\text{b}}=\vec{\text{c}}$
- $\vec{\text{b}}+\vec{\text{c}}=\vec{\text{a}}$
View full question & answer→Question 263 Marks
If D is the mid-point of side BC of a triangle ABC such that $\overrightarrow{\text{AB}}+\overrightarrow{\text{AC}}=\lambda\overrightarrow{\text{AD}}$, write the value of $\lambda$.
AnswerGiven: D is the mid-point of the side BC of a triangle ABC such that $\overrightarrow{\text{AB}}+\overrightarrow{\text{AC}}=\lambda\overrightarrow{\text{AD}}$Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are the position vectors of AB, BC and CA.
Now, the position vector of D is $\frac{\vec{\text{b}}+\vec{\text{c}}}2$. Then,
$\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$
$\overrightarrow{\text{AC}}=\vec{\text{c}}-\vec{\text{a}}$
$\overrightarrow{\text{AD}}=\frac{\vec{\text{b}}+\vec{\text{c}}}2-\vec{\text{a}}$
Now, we have
$\overrightarrow{\text{AB}}+\overrightarrow{\text{AC}}=\lambda\overrightarrow{\text{AD}}$
$\Rightarrow\ \vec{\text{b}}-\vec{\text{a}}+\vec{\text{c}}-\vec{\text{a}}=\lambda\Big(\frac{\vec{\text{b}}+\vec{\text{c}}-\vec{\text{a}}}2\Big)$
$\Rightarrow\ \vec{\text{b}}+\vec{\text{c}}-2\vec{\text{a}}=\lambda\Big(\frac{\vec{\text{b}}+\vec{\text{c}}-2\vec{\text{a}}}2\Big)$
$\Rightarrow\ \lambda=2$
View full question & answer→Question 273 Marks
ABCD is a quadrilateral. Find the sum of the vectors $\overrightarrow{\text{BA}},\overrightarrow{\text{BC}},\overrightarrow{\text{CD}}\text{ and }\overrightarrow{\text{DA}}$.
AnswerGiven: ABCD is a quadrilateral.
We need to find the sum of $\overrightarrow{\text{BA}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DA}}$.
Consider,
$\overrightarrow{\text{BA}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DA}}$
$=\Big(\overrightarrow{\text{BA}}+\overrightarrow{\text{DA}}\Big)+\Big(\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}\Big)$
$=\Big(\overrightarrow{\text{BD}}+2\overrightarrow{\text{DA}}\Big)+\overrightarrow{\text{BD}}$ $\Big[\because\ \overrightarrow{\text{BD}}+\overrightarrow{\text{DA}}=\overrightarrow{\text{BA}}\text{ and }\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{BD}}\Big]$
$=2\Big(\overrightarrow{\text{BD}}+\overrightarrow{\text{DA}}\Big)$
$=2\ \overrightarrow{\text{BA}}$
View full question & answer→Question 283 Marks
ABCDE is a pentagon, prove that,$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$
AnswerGiven ABCDE is a pentagon.
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$
$=\Big(\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}\Big)+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$ $\Big[$Using triangle law in $\triangle\text{ABC},\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\Big]$
$=\Big(\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}\Big)+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$
$=\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$ $\Big[$Using triangle law in $\triangle\text{ACD},\ \overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}\Big]$
$=\overrightarrow{\text{AD}}+\overrightarrow{\text{DA}}$
$=\overrightarrow{\text{AD}}-\Big(-\overrightarrow{\text{AD}}\Big)$
$=\vec0$
$\therefore\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$
View full question & answer→Question 293 Marks
Express $\overrightarrow{\text{AB}}$ in terms of unit vectors $\hat{\text{i}}\text{ and }\hat{\text{j}}$, when the point is:A(-6, 3), B(-2, -5)
Find $\Big|\overrightarrow{\text{AB}}\Big|$
AnswerHere, A = (-6, 3) B = (-2, -5) Position vector of $\text{A}=-6\hat{\text{i}}+3\hat{\text{j}}$ Position vector of $\text{B}=-2\hat{\text{i}}-5\hat{\text{j}}$ $\overrightarrow{\text{AB}}$ = Position vector of B - Position vector of A$=\big(-2\hat{\text{i}}-5\hat{\text{j}}\big)-\big(-6\hat{\text{i}}+3\hat{\text{j}}\big)$
$=-2\hat{\text{i}}-5\hat{\text{j}}+6\hat{\text{i}}-3\hat{\text{j}}$
$\overrightarrow{\text{AB}}=4\hat{\text{i}}-8\hat{\text{j}}$
$\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(4)^2+(-8)^2}$
$=\sqrt{16+64}$
$=\sqrt{80}$
$=\sqrt{16\times5}$ $=4\sqrt5$$\Big|\overrightarrow{\text{AB}}\Big|=4\sqrt5$
$\overrightarrow{\text{AB}}=4\hat{\text{i}}-8\hat{\text{j}}$
View full question & answer→Question 303 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\pm\vec{\text{b}}$
AnswerFalse
$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\pm\vec{\text{b}}$
Consider an example,
$\vec{\text{a}}=\text{i}+\sqrt3\text{j}$ and $\vec{\text{b}}=\sqrt2\text{i}+\sqrt2\text{j}$
$\big|\vec{\text{a}}\big|=\sqrt{1^2+\big(\sqrt3\big)^2}=2$ and $\big|\vec{\text{b}}\big|=\sqrt{\big(\sqrt2\big)^2+\big(\sqrt2\big)^2}=2$
Thus, $\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$ but $\vec{\text{a}}\neq\pm\vec{\text{b}}$
View full question & answer→Question 313 Marks
Express $\overrightarrow{\text{AB}}$ in terms of unit vectors $\hat{\text{i}}\text{ and }\hat{\text{j}}$, when the point is:A(4, -1), B(1, 3)
Find $\Big|\overrightarrow{\text{AB}}\Big|$
AnswerHere, A = (4, -1) B = (1, 3) Position vector of $\text{A}=4\hat{\text{i}}-\hat{\text{j}}$ Position vector of $\text{B}=\hat{\text{i}}+3\hat{\text{j}}$ $\overrightarrow{\text{AB}}$ = Position vector of B - Position vector of A$=\big(\hat{\text{i}}+3\hat{\text{j}}\big)-\big(4\hat{\text{i}}-\hat{\text{j}}\big)$
$=\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{i}}+\hat{\text{j}}$
$\overrightarrow{\text{AB}}=-3\hat{\text{i}}+4\hat{\text{j}}$
$\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(-3)^2+(4)^2}$
$=\sqrt{9+16}$
$=\sqrt{25}$
$\Big|\overrightarrow{\text{AB}}\Big|=5$
$\overrightarrow{\text{AB}}=-3\hat{\text{i}}+4\hat{\text{j}}$
View full question & answer→Question 323 Marks
Find the unit vector in the direction of the resultant of the vectors $\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}},\ 2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$.
AnswerGiven: $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}},\ 2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$ are the position vectors. Then, Resultant of vectors $=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$$=\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}+2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}+\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
$=4\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
So, $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{4^2+2^2+1^2}$$=\sqrt{16+4+1}$
$=\sqrt{21}$
$\therefore$ Unit vector in the direction of the resultant vector $=\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}{\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|}$ $=\frac{1}{\sqrt{21}}\big(4\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)$
View full question & answer→Question 333 Marks
A vector makes an angle of $\frac{\pi}4$ with each of x-axis and y-axis. Find the angle made by it with the z-axis.
AnswerLet the vector $\overrightarrow{\text{OP}}$ makes an angle $\alpha=45^{\circ}$ and $\beta=45^{\circ}$ with OX, OY respectively. Suppose $\overrightarrow{\text{OP}}$ is inclined at angle $\gamma$ to OZ.
Let l, m, n be the direction cosines of $\overrightarrow{\text{OP}}$. Then,
$\text{l}=\cos\frac{\pi}4=\frac{1}{\sqrt2}$
$\text{m}=\cos\frac{\pi}4=\frac{1}{\sqrt2}$
$\text{n}=\cos\gamma$
Now, we have,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \frac{1}2+\frac{1}2+\text{n}^2=1$
$\Rightarrow\ \text{n}^2=0$
$\Rightarrow\ \text{n}=0$
$\Rightarrow\ \cos\gamma=\cos\frac{\pi}2$
$\Rightarrow\ \gamma= \frac{\pi}2$
Hence, the angle made by it with the z-axis is $\frac{\pi}2$.
View full question & answer→Question 343 Marks
Using vectors, find the value of $\lambda$ such that the points ($\lambda$, -10, 3), (1, -1, 3) and (3, 5, 3) are collinear.
AnswerPoints ($\lambda$, -10, 3), (1, -1, 3) and (3, 5, 3) are collinear.$\therefore$ ($\lambda$, -10, 3) = x(1, -1, 3) + y(3, 5, 3) for some scalars x and y.
$\Rightarrow\lambda$ = x + 3y, -10 = -x + 5y and 3 = 3x +3y
Solving -10 = -x + 5y and 3 = 3x + 3y for x and y we get,
$\text{x}=\frac{5}2$ and $\text{y}=-\frac{3}2$
Now,
$\lambda=\text{x}+3\text{y}$
$\Rightarrow\lambda=\frac{5}2+3\Big(-\frac{3}2\Big)=-2$
View full question & answer→Question 353 Marks
If the position vectors of the points A(3, 4), B(5, -6) and C(4, -1) are $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ respectively, compute $\vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}$.
AnswerHere, A(3, 4), B(5, -6), C(4, -1) $\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}$ $\vec{\text{b}}=5\hat{\text{i}}-6\hat{\text{j}}$ $\vec{\text{c}}=4\hat{\text{i}}-\hat{\text{j}}$ Now,$\vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}=\big(3\hat{\text{i}}+4\hat{\text{j}}\big)+2\big(5\hat{\text{i}}-6\hat{\text{j}}\big)-3\big(4\hat{\text{i}}-\hat{\text{j}}\big)$
$=3\hat{\text{i}}+4\hat{\text{j}}+10\hat{\text{i}}-12\hat{\text{j}}-12\hat{\text{i}}+3\hat{\text{j}}$ $=\hat{\text{i}}-5\hat{\text{j}}$ $\therefore\ \vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}=\hat{\text{i}}-5\hat{\text{j}}$
View full question & answer→Question 363 Marks
Using vector method, prove that the point is collinear:
A(2, -1, 3), B(4, 3, 1) and C(3, 1, 2)
AnswerGiven the points A(2, -1, 3), B(4, 3, 1) and C(3, 1, 2). Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=4\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}-2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}$
$=-2\big(-\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}-4\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}$
$=-\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=-2\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given points A, B, and C are collinear.
View full question & answer→Question 373 Marks
Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}},\vec{\text{d}}$ be the position vectors of the four distinct points A, B, C, D. If $\vec{\text{b}}-\vec{\text{a}}=\vec{\text{c}}-\vec{\text{d}}$, then show that ABCD is a parallelogram.
AnswerHere it is given that $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}\text{ and }\vec{\text{d}}$ be the position vectors of the four distinct points A, B, C, D such that,
$\vec{\text{b}}-\vec{\text{a}}=\vec{\text{c}}-\vec{\text{d}}$
Given that,
$\vec{\text{b}}-\vec{\text{a}}=\vec{\text{c}}-\vec{\text{d}}$
$\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}$
So, AB is parallel and equal to DC (in magnitude).
Hence,
ABCD is a parallelogram.
View full question & answer→Question 383 Marks
Using vector method, prove that the point is collinear:
A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1)
AnswerGiven the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1). Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-7\hat{\text{k}}$
$=\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=3\hat{\text{i}}+10\hat{\text{j}}-\hat{\text{k}}-2\hat{\text{i}}-6\hat{\text{j}}-3\hat{\text{k}}$
$=\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given points A, B, and C are collinear.
View full question & answer→