Question 13 Marks
Find the vector and cartesian equations of the planes:
That passes through the point (1, 4, 6) and the normal vector to the plane is $\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}.$
AnswerThe position vector of point (1, 4, 6) is $\vec{\text{a}}=\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}}$
The normal vector $\vec{\text{N}}$ perpendicular to the plane is $\vec{\text{N}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
The vector equation of the plane is given by, $\Big(\vec{\text{r}}-\vec{\text{a}}\Big).\vec{\text{N}}=0$
$\Rightarrow\Big[\vec{\text{r}}-\Big(\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}}\Big)\Big].\Big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\Big)=0\ \ \ ....(1)$
$\vec{\text{r}}$ is the position vector of any point P(x, y, z) in the plane.
$\therefore\ \ \vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
Therefore, equation (1) becomes
$\Big[\Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\Big)-\Big(\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}}\Big)\Big].\Big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\Big)=0$
$\Rightarrow\Big[(\text{x}-1)\hat{\text{i}}+(\text{y}-4)\hat{\text{j}}+(\text{z}-6)\hat{\text{k}}\Big].\Big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\Big)=0$
⇒ (x - 1) - 2(y - 4) + (z - 6) = 0
⇒ x - 2y + z +1 = 0
This is the Cartesian equation of the required plane.
View full question & answer→Question 23 Marks
Find the direction cosines of the line passing through two points $(-2, 4, -5)$ and $(1, 2, 3)$.
AnswerThe direction consines of the line passing through two points $P x_1, y_1, z_1$, and $Q (x_2, y_2, z_2)$ are $\frac{\text{x}_2-\text{x}_1}{\text{PQ}},\frac{\text{y}_2-\text{y}_1}{\text{PQ}},\frac{\text{z}_2-\text{z}_1}{\text{PQ}}.$
Here,
$\text{PQ}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2+(\text{z}_2-\text{z}_1)^2}$
$\text{P}=2,4,-5$
$\text{Q}=1,2,3$
$\therefore\text{PQ}=1-(-2)^2+(2-4)^2+[3-(-5)]^2=\sqrt{77}$
Thus, the direction cosines of the line joining two points are
$\frac{1-(-2)}{\sqrt{77}},\frac{2-4}{\sqrt{77}},\frac{3-(-5)}{\sqrt{77}},\text{i.e.}\frac{3}{\sqrt{77}}77,\frac{-2}{\sqrt{77}}77,\frac{8}{\sqrt{77}}.$
View full question & answer→Question 33 Marks
Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=6$ and $\vec{\text{r}}\cdot(3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}})=9$
AnswerWe know that the angle between the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ is given by
$\cos\theta=\frac{\vec{\text{n}}_1\cdot\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{n}}_2=3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}}$
So, $\cos\theta=\frac{(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})\cdot(3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}})}{\big|2\hat{\text{i}}-3\hat{\text{j}}+4{\hat{\text{k}}\big|}\big|3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{k}}\big|}$
$=\frac{6-6-4}{\sqrt{4+1+4}\sqrt{9+36+4}}$
$=\frac{-4}{(3)(7)}$
$=\frac{-4}{21}$
$\theta=\cos^{-1}\Big(\frac{-4}{21}\Big)$
View full question & answer→Question 43 Marks
If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (-4, 3, -6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
AnswerThe coordinates of A, B, C, and D are (1, 2, 3), (4, 5, 7), (-4, 3, -6), and (2, 9, 2) respectively.
The direction ratios of AB are (4 - 1) = 3, (5 - 2) = 3, and (7 - 3) = 4
The direction ratios of CD are (2 - (-4)) = 6, (9 - 3) = 6, and (2 - (-6)) = 8
It can be seen that, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}=\frac{1}{2}$
Therefore, AB is parallel to CD.
Thus, the angle between AB and CD is either 0° or 180°.
View full question & answer→Question 53 Marks
Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′, respectively, from the origin, prove that $\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}=\frac{1}{\text{a}'^2}+\frac{1}{\text{b}'^2}+\frac{1}{\text{c}'^2}.$
AnswerConsider OX, OY, OZ and ox, oy, oz are two system of rectangular axes. Let their corresponding equation of plane be
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ....(\text{i})$
And $\frac{\text{x}}{\text{a}'}+\frac{\text{y}}{\text{b}'}+\frac{\text{z}}{\text{c}'}=1\ ....(\text{ii})$
Also the length of perpendicular from origin to equations (i) and (ii) must be same.
$\therefore\frac{\frac{0}{\text{a}}+\frac{0}{\text{b}}+\frac{0}{\text{c}}-1}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}}=\frac{\frac{0}{\text{a}'}+\frac{0}{\text{b}'}+\frac{0}{\text{c}'}-1}{\sqrt{\frac{1}{\text{a}'^2}+\frac{1}{\text{b}'^2}+\frac{1}{\text{c}'^2}}}$
$\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}=\frac{1}{\text{a}'^2}+\frac{1}{\text{b}'^2}+\frac{1}{\text{c}'^2}$
View full question & answer→Question 63 Marks
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\vec{\text{r}}.\Big(\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}\Big)+9=0.$
AnswerThe required line passes through the point P(1, 2, 3).
$\therefore$ Position vector $\vec{\text{a}}$ (say) of point P is (1, 2, 3)
$\Rightarrow\ \ \vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Equation of the given plane is $\vec{\text{r}}.\Big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\Big)+9=0$
$\vec{\text{r}}.\Big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\Big)=-9$
Comparing with $\vec{\text{r}}.\vec{\text{n}}=\vec{\text{d}},\ \ \ \vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}$
Since, the required line is perpendicular to the given plane, therefore, vector $\vec{\text{b}}$ along the required line is $\vec{\text{b}}=\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}+-5\hat{\text{k}}$
$\because$ Equation of the required line is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\therefore\ \ \vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}\Big)$
View full question & answer→Question 73 Marks
The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.
Answer$\text{x}=\text{ay+b},$
$\text{z}=\text{cy+d}$
$\frac{\text{x}-\text{b}}{\text{a}}=\frac{\text{y}}{1}=\frac{\text{z}-\text{d}}{\text{c}}=\lambda$ (say)
So DR's of line are (a, 1, c)
from above equation, we can write
$\text{x}=\text{a}\lambda+\text{b}$
$\text{y}=\lambda$
$\text{z}=\text{c}\lambda+\text{d}$
So vector equation of line is
$\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=(\text{b}\hat{\text{i}}+\text{d}\hat{\text{k}})+\lambda(\text{a}\hat{\text{i}}+\text{x}\hat{\text{j}}+\text{c}\hat{\text{k}}\big)$
View full question & answer→Question 83 Marks
Find the distance of the plane 2x- 3y + 4z - 6 = 0 from the origin.
AnswerThe given equation of the plane is,
2x - 3y + 4z = 6 .....(i)
Now, $\sqrt{2^2+(-3)^2+4^2}$
$=\sqrt{4+9+16}$
$=\sqrt{29}$
Dividing (i) by $\sqrt{29},$ we get
$\frac{2}{\sqrt{29}}\text{x}-\frac{3}{\sqrt{29}}\text{y}+\frac{4}{\sqrt{29}}\text{z}=\frac{6}{\sqrt{29}},$ which is the normal form of plane (i)
So, the length of the perpendicular from the origin to the plane $=\frac{6}{\sqrt{29}}$
View full question & answer→Question 93 Marks
Write the direction cosines of the line $\frac{\text{x}-2}{2}=\frac{2\text{y}-5}{-3},\text{z}=2.$
AnswerWe have$\frac{\text{x}-2}{2}=\frac{2\text{y}-5}{-3},\text{z}=2$
The equation of given line can be re-written as
$\frac{\text{x}-2}{2}=\frac{\text{y}-\frac{5}{2}}{-\frac{3}{2}}=\frac{\text{z}-2}{0}$
$\frac{\text{x}-2}{4}=\frac{\text{y}-\frac{5}{2}}{-3}=\frac{\text{z}-2}{0}$
The direction ratios of the given line are proportional to 4, -3, 0.
Hence, the direction cosines of the given line are proportional to
$\frac{4}{\sqrt{4^2+(-3)^2+0^2}},\frac{-3}{\sqrt{4^2+(-3)^2+0^2}},\frac{0}{\sqrt{4^2+(-3)^2+0^2}}$
$=\frac{4}{5},\frac{-3}{5},0$
View full question & answer→Question 103 Marks
Find the equation of the line passing through the points (2, 1, 3) and perpendicular to the lines $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}-3}{3}$ and $\frac{\text{x}}{-3}=\frac{\text{y}}{2}=\frac{\text{z}}{5}$
AnswerLet:
$\vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}_2=-3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}$
Since the required line is perpendicular to the lines parallel to the vectors
$\vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{b}}_2=-3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}},$ it is parallel to the vector $\vec{\text{b}}=\vec{\text{b}}_1\times\vec{\text{b}}_2.$
Now,
$\vec{\text{b}}=\vec{\text{b}}_1\times\vec{\text{b}}_2$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\-3&2&5\end{vmatrix}$
$=4\hat{\text{i}}-14\hat{\text{j}}+8\hat{\text{k}}$
$=2\big(2\hat{\text{i}}-7\hat{\text{j}}+4\hat{\text{k}}\big)$
Thus, the diraction ratios of the required line are proportional to 2, -7, 4.
The equation of the required line passing through the point (2, 1, 3) and having direction ratios proportional to 2, -7, 4 is $\frac{\text{x}-2}{2}=\frac{\text{y}-1}{-7}=\frac{\text{z}-3}{4}.$
View full question & answer→Question 113 Marks
Find the distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z - 9 = 0$.
AnswerWe know that the distance of the point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is given by
$\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|(2)+2(3)-2(-5)-9|}{\sqrt{1^2+2^2+(-2)^2}}$
$=\frac{|2+6+10-9|}{\sqrt{1+4+4}}$
$=\frac{9}{3}$
$=3\text{ units}$
View full question & answer→Question 123 Marks
In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.
$2x - y + 3z - 1 = 0$ and $2x - y + 3z + 3 = 0$
AnswerThe direction ratios of normal to the plane, $L_1: a_1x + b_1y + c_1z = 0$,
are $a_1, b_1, c_1$ and $L_2: a_2x + b_2y + c_2z = 0$ are $a_2, b_2, c_2$
$\text{L}_1||\text{L}_2,\ \text{if }\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\text{L}_1\perp\text{L}_2,\ \text{if}\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
The angle between $L_1$ and $L_2$ is given by,
$\text{Q}=\cos^{-1}\Bigg|\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}.\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}_2^2}}$
The equations of the planes are $2x - y + 3z - 1 = 0$ and $2x - y + 3z + 3 = 0$
Here, $a_1 = 2, b_1 = -1, c_1 = 3$ and $a_2 = 2, b_2 = -1, c_2 = 3$
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{2}=1,\ \frac{\text{b}_1}{\text{b}_2}=\frac{-1}{-1}=1\text{ and } \frac{\text{c}_1}{\text{c}_2}=\frac{3}{3}=1$
$\therefore\ \ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Thus, the given lines are parallel to each other.
View full question & answer→Question 133 Marks
Find the equation of the plane through the untersection of the planes 3x - y + 2z = 4 and x + y + z = 2 and the point (2, 2, 1).
AnswerThe equation of the family of planes through the intersection of planes 3x - y + 2z = 4 and x + y + z = 2 is,
$(3\text{x}-\text{y}+2\text{z}-4)+\lambda(\text{x}+\text{y}+\text{z}-2)=0\ ....(\text{i})$
If it passes through (2, 2, 1), then
$(6-2+2-4)+\lambda(2+2+1-2)=0$
$\lambda=-\frac{2}{3}$
Substituting $\lambda=-\frac{2}{3}$ in (i) we get, 7x - 5y + 4z = 0 as the equation of the required plane.
View full question & answer→Question 143 Marks
Reduce the equation of the following planes to intercept from and find the intercepts on the coordinate axes:
4x + 3y - 6z - 12 = 0
AnswerEquation of the given plane is,
4x + 3y - 6z - 12 = 0
⇒ 4x + 3y - 6z = 12
Dividng both sides by 12, we get
$\frac{4\text{x}}{12}+\frac{3\text{y}}{12}+\frac{(6\text{z})}{12}=\frac{12}{12}$
$\Rightarrow\frac{4\text{x}}{12}+\frac{3\text{y}}{12}-\frac{6\text{z}}{12}=\frac{12}{12}$
$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{4}+\frac{\text{z}}{-2}=1\ ...(\text{i})$
We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{ii})$
Comparing (i) and (ii) we get
$\text{a}=3;\text{ b}=4;\text{ c}=-2$
View full question & answer→Question 153 Marks
Determine the value of $\lambda$ for which the following planes are perpendicular to other.
$3\text{x}-6\text{y}-2\text{z}=7$ and $2\text{x}+\text{y}-\lambda\text{z}=5$
AnswerWe know that the planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perperndicular to each other only if
$a_1a_2 + b_1b_2 + c_1c_2= 0$
The given planes are $3x - 6y - 2z = 7 $and $2\text{x}+\text{y}-\lambda\text{z}=5$
$\Rightarrow a_1 = 3; b_1 = -6; c_1 = -2; a_2 = 2; b_2 = 1$; $\text{c}_2=-\lambda$
It is given that the given planes are perpendicular.
$\Rightarrow a_1a_2 + b_1b_2 + c_1c_2 = 0$
$\Rightarrow(3)(2)+(-6)(1)+(-2)(-\lambda)=0$
$\Rightarrow6-6+2\lambda=0$
$\Rightarrow2\lambda=0$
$\Rightarrow\lambda=0$
View full question & answer→Question 163 Marks
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the:
yz-plane
AnswerDirection ratios of the given line are
(5 - 3, 1 - 4, 6 - 1) = (2, -3, 5)
Hence, equation of the line is
$\frac{\text{x}-5}{2}=\frac{\text{y}-1}{-3}=\frac{\text{z}-6}{5}=\text{r}$
$\Rightarrow\text{x}=2\text{r}+5,\text{ y}=-3\text{r}+1,\text{ z}=5\text{r}+6$
For any point on the yz-plane x = 0
$\Rightarrow2\text{r}+5=0$
$\Rightarrow\text{r}=-\frac{5}{2}$
$\Rightarrow\text{y}=-3\Big(-\frac{5}{2}\Big)+1=\frac{17}{2}$
$\Rightarrow\text{z}=5\Big(-\frac{5}{2}\Big)+6=-\frac{13}{2}$
Hence, the corrdinates of the point are $\Big(0,\frac{17}{2},-\frac{13}{2}\Big)$
View full question & answer→Question 173 Marks
Find the angle between the plane:
x + y - 2z = 3 and 2x - 2y + z = 5
AnswerWe know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between x + y - 2z = 3 and 2x - 2y + z = 5 is given by
$\cos\theta=\frac{(1)(2)+(1)(-2)+(-2)(1)}{\sqrt{1^2+1^2+(-2)^2}\sqrt{2^2+(-2)^2+1^2}}$
$=\frac{2-2-2}{\sqrt{1+1+4}\sqrt{4+4+1}}=\frac{-2}{\sqrt{6}\sqrt{9}}$
$=\frac{-2}{\sqrt{6}\sqrt{9}}=\frac{-2}{3\sqrt{6}}$
$\theta=\cos^{-1}\Big(\frac{-2}{3\sqrt{6}}\Big)$
View full question & answer→Question 183 Marks
Find the angle between the line $\frac{\text{x}+1}{2}=\frac{\text{y}}{3}=\frac{\text{z}-3}{6}$ and the plane 10x + 2y - 11z = 3.
AnswerThe given line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$ and the given plane is normal to the vector $\vec{\text{n}}=10\hat{\text{i}}+2\hat{\text{j}}-11\hat{\text{k}}$
We know that the angle $\theta$ between the line and the plane is given by.
$\sin\theta=\frac{\vec{\text{b}}\cdot\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})\cdot(10\hat{\text{i}}+2\hat{\text{j}}-11\hat{\text{k}})}{|2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}||10\hat{\text{i}}+2\hat{\text{j}}-11\hat{\text{k}}|}$
$=\frac{20+6-66}{\sqrt{4+9+36}\sqrt{100+4+121}}$
$=\frac{-40}{(7)(15)}=\frac{-8}{21}$
$\theta=\sin^{-1}\Big(\frac{-8}{21}\Big)$
View full question & answer→Question 193 Marks
Find the coordinates of the point where the line through (3, -4, -5) and (2, -3, 1) corsses the plane 2x + y + z = 7.
AnswerThe equation of the through the points (3, -4, -5) and (2, -3, 1) is
$\frac{\text{x}-3}{2-3}=\frac{\text{y}+4}{-3+4}=\frac{\text{z}+5}{1+5}$
$\Rightarrow\frac{\text{x}-3}{-1}=\frac{\text{y}+4}{1}=\frac{\text{z}+5}{6}$
The coordinates of any point on this line are of the form
$\frac{\text{x}-3}{-1}=\frac{\text{y}+4}{1}=\frac{\text{z}+5}{6}=\lambda$
$\Rightarrow\text{x}=-\lambda+3,\text{ y}=\lambda-4,\text{ z}=6\lambda-5$
$\Rightarrow5\lambda=10$
$\Rightarrow\lambda=2$
So, the coordinates of the point are
$(-\lambda+3,\lambda-4,6\lambda-5)$
$=(-2+3,2-4,6(2)-5)$
$=(1,-2,7)$
View full question & answer→Question 203 Marks
Find the angle between the following pairs of lines:
-
$\vec{\text{r}}=2\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}+\lambda\Big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\Big)\ \text{and}$
$\vec{\text{r}}=7\hat{\text{i}}-6\hat{\text{k}}+\mu\Big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\Big)$
Answer
- Equation of the first line is $\vec{\text{r}}=2\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}+\lambda\Big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\Big)$
Comparing with $\Big(\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\Big),$
$ \vec{\text{a}_1}=2\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\ \text{and}\ \vec{\text{b}_1}=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
(Vector $\vec{\text{a}}$ is the position vector of a point on line $\vec{\text{b}}$ is a vector along the line)
Again, equation of the second line is $\vec{\text{r}}=7\hat{\text{i}}-6\hat{\text{k}}+\mu\Big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\Big)$
Comparing with $\Big(\vec{\text{r}}=\vec{\text{a}}+\mu\vec{\text{b}}\Big),$
$\vec{\text{a}_2}=7\hat{\text{i}}-6\hat{\text{k}}\ \text{and}\ \vec{\text{b}_2}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
(Vector $\vec{\text{a}}$ is the position vector of a point on line $\vec{\text{b}}$ is a vector along the line)
Let $\theta$ be the angle between these two lines, then
$\cos\theta=\frac{\vec{\text{b}_1}.\vec{\text{b}_2}}{\Big|\vec{\text{b}_1}\Big|.\Big|\vec{\text{b}_2}\Big|} =\frac{3(1)+2(2)+6(2)}{\sqrt{9+4+36}\sqrt{1+4+4}}=\frac{3+4+12}{\sqrt{49}\sqrt{9}}=\frac{19}{7\times3}$
$\cos\theta=\frac{19}{21}\ \ \ \ \ \ \ \Rightarrow\ \theta=\cos^{-1}\frac{19}{21}$ View full question & answer→Question 213 Marks
Cartesian equations of a line AB are $\frac{2\text{x}-1}{2}=\frac{4-\text{y}}{7}=\frac{\text{z}+1}{2}.$ Write the direction ratios of a parallel to AB.
Answer$\frac{2\text{x}-1}{2}=\frac{4-\text{y}}{7}=\frac{\text{z}+1}{2}$The equation of the line AB can be re-written as
$\frac{\text{x}-\frac{1}{2}}{1}=\frac{\text{y}-4}{-7}=\frac{\text{z}+1}{2}$
The direction ratios of the line parallel to AB are proportional to 1, -7, 2.
Also, the diraction cosines of the line parallel to AB are proportional to
$\frac{1}{\sqrt{1^2+(-7)^2+2^2}},\frac{-7}{\sqrt{1^2+(-7)^2+2^2}},\frac{2}{\sqrt{1^2+(-7)^2+2^2}}$
$=\frac{1}{\sqrt{54}},\frac{-7}{\sqrt{54}},\frac{2}{\sqrt{54}}$
View full question & answer→Question 223 Marks
Find the equation of the plane through the intersection of the planes 3x - y + 2z - 4 = 0 and x + y + z - 2 = 0 and the point (2, 2, 1).
AnswerThe equation of any plane through the intersection of the planes,
3x - y + 2z - 4 = 0 and x + y + z - 2 = 0, is
$(3\text{x}-\text{y}+2\text{z}-4)+\alpha(\text{x}+\text{y}+\text{z}-2)=0,\text{ where }\alpha\in\text{R}\ \ ....(1)$
The plane passes through the point (2, 2, 1).
Therefore, this point will satisfy equation (1)
$\therefore\ (3\times2-2+2\times1-4)+\alpha(2+2+1-2)=0$
$\Rightarrow\ 2+3\alpha=0$
$\Rightarrow\ \alpha=-\frac{2}{3}$
Substituting $\alpha=-\frac{2}{3}$ in equation (1), we obtain
$(3\text{x}-\text{y}+2\text{z}-4)-\frac{2}{3}(\text{x}+\text{y}+\text{z}-2)=0,$
⇒ 3(3x - y + 2z - 4) - 2(x + y + z - 2) = 0
⇒ (9x - 3y + 6z - 12) - 2(x + y + z - 2) = 0
⇒ 7x - 5y + 4z - 8 = 0
This is the required equation of the plane.
View full question & answer→Question 233 Marks
Find the distance of the point $(2, 3, 5)$ from the xy-plane.
AnswerWe know that, the distance (D) of a the point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is given by
$\text{D}=\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}\ ...(\text{i})$
So, distance of point (2, 3, 5) from xy-plane (we know that equation of xy-plane is z = 0) is
$=\Bigg|\frac{(2)(0)+(3)(0)+(5)(1)+0}{\sqrt{(0)^2+(0)^2+(1)^2}}\Bigg|$ [Using (i)]
$=\frac{0+0+5}{\sqrt{0+0+1}}$
$=5\text{ unit}$
Distance of the point (2, 3, 5) from xy-plane = 5 unit
View full question & answer→Question 243 Marks
Find the angle between the following pair of lines:
- $\frac{\text{x}}{2}=\frac{\text{y}}{2}=\frac{\text{z}}{1}\ \text{and}\ \frac{\text{x}-5}{4}=\frac{\text{y}-2}{1}=\frac{\text{z}-3}{8}$
Answer
- Given: Equation of first line is $\frac{\text{x}}{2}=\frac{\text{y}}{2}=\frac{\text{z}}{1}$
The direction ratios of this line i.e., a vector along the line is
$\vec{\text{b}_1}=(2,\ 2,\ 1)=2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
Now equation of second line is $\frac{\text{x}-5}{4}=\frac{\text{y}-2}{1}=\frac{\text{z}-3}{8}$
The direction ratios of this line i.e., a vector along the line is
$\vec{\text{b}_2}=(4,\ 1,\ 8)=4\hat{\text{i}}+\hat{\text{j}}+8\hat{\text{k}}$
Let $\theta$ be the angle between these two lines, then
$\cos\theta=\frac{\vec{\text{b}_1}.\vec{\text{b}_2}}{\Big|\vec{\text{b}_1}\Big|.\Big|\vec{\text{b}_2}\Big|} =\frac{2(4)+(2)(1)+(1)(8)}{\sqrt{4+4+1}\sqrt{16+1+64}}=\frac{8+2+8}{\sqrt{9}\sqrt{81}}=\frac{18}{3\times9}=\frac{2}{3}$
$\Rightarrow\ \ \ \theta=\cos^{-1}\frac{2}{3}$ View full question & answer→Question 253 Marks
Find the coordinates of the point where the line through (3, -4, -5) and (2, -3, 1) crosses the plane 2x + y + z = 7.
AnswerDirection ratios of the line joining the points A(3, -4, -5) and B(2, -3, 1) are 2 - 3, -3 - (-4), 1 - (-5) ⇒ -1, 1, 6
$\therefore$ Equation of the line AB are
$\frac{\text{x}-3}{-1}=\frac{\text{y}+4}{1}=\frac{\text{z}+5}{6}\ \ \ .....(\text{i})$
Equation of the plane is 2x + y + z = 7 ...(ii)
Now to find the point where line (i) crosses plane (ii),
From eq. (i) $\frac{\text{x}-3}{-1}=\frac{\text{y}+4}{1}=\frac{\text{z}+5}{6}=\lambda\ (\text{say})$
$\Rightarrow\ \ \text{x}-3=-\lambda,\ \text{y}+4=\lambda,\ \text{z}+5=6\lambda$
$\Rightarrow\ \ \text{x}=3-\lambda,\ \text{y}=-4+\lambda,\ \text{z}=-5+6\lambda\ \ \ ....(\text{iii})$
Putting the values of x, y, z in eq. (ii), we get
$2(3-\lambda)+(-4+\lambda)+(-5+6\lambda)=7$
$\Rightarrow\ \ 6-2\lambda-4+\lambda-5+6\lambda=7\ \ \Rightarrow\ 5\lambda=10\ \ \Rightarrow\ \lambda=2$
Putting $\lambda=2$ in eq. (iii), point of intersection of line (i) and plane (ii) is
x = 3 - 2 = 1, y = -4 + 2 = -2, z = -5 + 12 = 7
Thus, required point of intersection is (1, -2, 7).
View full question & answer→Question 263 Marks
Reduce the equation of the following planes to intercept from and find the intercepts on the coordinate axes:
2x + 3y - z = 6
AnswerThe equation of the given plane is,2x + 3y - z = 6
Dividng both sides by 6, we get
$\frac{2\text{x}}{6}+\frac{3\text{y}}{6}-\frac{\text{z}}{6}=\frac{6}{6}$
$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{2}+\frac{\text{z}}{-6}=1\ ...(\text{i})$
We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{ii})$
Comparing (i) and (ii) we get
$\text{a}=3;\text{ b}=2;\text{ c}=-6$
View full question & answer→Question 273 Marks
Find the angle between the line $\vec{\text{r}}=(2\hat{\text{i}}+3\hat{\text{j}}+9\hat{\text{k}})+\lambda(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})$ and the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=5$
AnswerWe know that the angle $\theta$ between the line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ and the plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$ is given by
$\sin\theta=\frac{\vec{\text{b}}\cdot\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
Here,
$\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
So, $\sin\theta=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})}{|2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}||\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}|}$
$=\frac{2+3+4}{\sqrt{4+9+16}\sqrt{1+1+1}}$
$=\frac{9}{\sqrt{29}\sqrt{3}}=\frac{3\sqrt{3}}{\sqrt{29}}$
$\theta=\sin^{-1}\Big(\frac{3\sqrt{3}}{\sqrt{29}}\Big)$
View full question & answer→Question 283 Marks
Find the angle between the plane:
2x + y - 2z = 5 and 3x - 6y - 2z = 7
AnswerWe know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between 2x + y - 2z = 5 and 3x - 6y - 2z = 7 is given by
$\cos\theta=\frac{(2)(3)+(1)(-6)+(-2)(-2)}{\sqrt{2^2+1^2+(-2)^2}\sqrt{3^2+(-6)^2+(-2)^2}}$
$=\frac{6-6+4}{\sqrt{4+1+4}\sqrt{9+36+4}}$
$=\frac{4}{(3)(7)}=\frac{4}{21}$
$\theta=\cos^{-1}\Big(\frac{4}{21}\Big)$
View full question & answer→Question 293 Marks
Determine the value of $\lambda$ for which the following planes are perpendicular to other.
$2\text{x}-4\text{y}+3\text{z}=5$ and $\text{x}+2\text{y}+\lambda\text{z}=5$
AnswerWe know that the planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perperndicular to each other only if
$a_1a_2 + b_1b_2 + c_1c_2= 0$
The given planes are $2x - 4y + 3z = 5$ and $\text{x}+2\text{y}+\lambda\text{z}=5$
$\Rightarrow a_1 = 2; b_1 = -4; c_1 = 3; a_2 = 1; b_2 = 2$; $\text{c}_2=\lambda$
It is given that the given planes are perpendicular.
$\Rightarrow a_1a_2 + b_1b_2 + c_1c_2 = 0$
$\Rightarrow(2)(1)+(-4)(2)+(3)+(\lambda)=0$
$\Rightarrow2-8+3\lambda=0$
$\Rightarrow3\lambda=6$
$\Rightarrow\lambda=2$
View full question & answer→Question 303 Marks
If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are $m_1n_2 - m_2n_1, l_1n_2 - l_2n_1, l_1m_2 - l_2m_1$.
Answer$l_1, m_1, n_1$ and $l_2, m_2, n_2$ are direction cosines of two mutually perpendicular of two given lines $L_1$ and $L_2$.(say)
Let $\hat{\text{n}}_1\ \text{and}\ \hat{\text{n}}_2$ be the unit vectors along these lines $L_1$ and $L_2$.
$\therefore\ \ \vec{\text{n}}_1=\text{l}_1\hat{\text{i}}+\text{m}_1\hat{\text{j}}+\text{n}_1\hat{\text{k}}\ \text{and}\ \vec{\text{n}}_2=\text{l}_2\hat{\text{i}}+\text{m}_2\hat{\text{j}}+\text{n}_2\hat{\text{k}}$
Let L be the line perpendicular to both the lines $L_1$ and $L_2$ and let $\hat{\text{n}}$ be a unit vector along line L perpendicular both lines $L_1$ and $L_2$.
$\therefore$ Cross product of two vectors
$=\hat{\text{n}}_1\times\hat{\text{n}}_2=\Big|\hat{\text{n}}_1\Big|.\Big|\hat{\text{n}}_2\Big|\sin90^{\circ}\hat{\text{n}}$ $\Big[\because\ \text{L}_1\perp\text{L}_2\ (\text{given},\ \therefore\ \text{angle between them is }90^{\circ}\Big]$
$\Rightarrow\ \ \hat{\text{n}}_1\times\hat{\text{n}}_2=\hat{\text{n}}$
$ \Rightarrow\ \hat{\text{n}}=\hat{\text{n}}_1\times\hat{\text{n}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\end{vmatrix}$
$\Rightarrow\ \ \hat{\text{n}}=(\text{m}_1\text{n}_2-\text{m}_2\text{n}_1)\hat{\text{i}}-(\text{l}_1\text{n}_2-\text{l}_2\text{n}_1)\hat{\text{j}}+(\text{l}_1\text{m}_2-\text{l}_2\text{m}_1)\hat{\text{k}}$
Since, $\hat{\text{n}}$ is a unit vector, therefore its components are its direction cosines.
Thus, direction cosines of $\hat{\text{n}}$ are $m_1n_2 - m_2n_1, l_1n_2 - l_2n_1, l_1m_2 - l_2m_1$
⇒ direction cosines of line L are $m_1n_2 - m_2n_1, l_1n_2 - l_2n_1, l_1m_2 - l_2m_1$.
View full question & answer→Question 313 Marks
Find the equation of the plane passing through the following points:
(2, 1, 0), (3, -2, -2) and (3, 1, 7)
AnswerThe equation of the plane passing through points (2, 1, 0), (3, -2, -2) and (3, 1, 7) is given by,$\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0\\3-2&-2-1&-2-0\\3-2&1-1&7-0\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0\\1&-3&-2\\1&0&7\end{vmatrix}=0$
$\Rightarrow-21(\text{x}-2)-9(\text{y}-1)+3\text{z}=0$
$\Rightarrow-21\text{x}+42-9\text{y}+9+3\text{z}=0$
$\Rightarrow-21\text{x}-9\text{y}+3\text{z}+51=0$
$\Rightarrow21\text{x}+9\text{y}-3\text{z}=51$
$\Rightarrow7\text{x}+3\text{y}-\text{z}=17$
View full question & answer→Question 323 Marks
Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (-1, -2, 1) and, (1, 2, 5).
AnswerThe diraction ratios of a line passing through the points (4, 7, 8) and (2, 3, 4) are
(4 - 2, 7 - 3, 8 - 4)
= (2, 4, 4)
The direction ratios of a line passing through the points
(-1, -2, 1) and (1, 2, 5)are
(-1 -1, -2 -2, 1 - 5)
= (-2, -4, -4)
The direction ratios are proportional.
$\frac{2}{-2}=\frac{4}{-4}=\frac{4}{-4}$
Hence, the lines are mutually parallel.
View full question & answer→Question 333 Marks
Find the vector equation of the line which is parallel to the vector $3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}$ and which passes throught the point (1, -2, 3).
AnswerLet $\vec{\text{a}}=3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}}$
So, vector equation of the line, which is parallel to the vector $\vec{\text{a}}=3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}}$ and passes through the vector $\vec{\text{b}}=\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}}$ is $\vec{\text{r}}=\vec{\text{b}}+\lambda\vec{\text{a}}.$
$\therefore\vec{\text{r}}=\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}}+\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}{\hat{\text{j}}}+\text{z}\hat{\text{k}})-(\hat{\text{i}}-2{\hat{\text{j}}}+3\hat{\text{k}})=\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$
$\Rightarrow(\text{x}-1)\hat{\text{i}}+(\text{y}+2)\hat{\text{j}}+(\text{z}-3)\hat{\text{k}}=\lambda(3\hat{\text{i}}-2{\hat{\text{j}}}+6\hat{\text{k}})$
View full question & answer→Question 343 Marks
Find the angle between the lines $2\text{x}=3\text{y}=-\text{z}$ and $6\text{x}=-\text{y}=-4\text{z}.$
AnswerThe equations of the given lines can be re-writen as
$\frac{\text{x}}{3}=\frac{\text{y}}{2}=\frac{\text{z}}{-6}$ and $\frac{\text{x}}{2}=\frac{\text{y}}{-12}=\frac{\text{z}}{-3}$
We know that angle between the lines $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$ is given by $\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}\sqrt{\text{a}^2_2+\text{b}^2_2+\text{c}^2_2}}.$
Let $\theta$ be the angle between the given lines.
$\therefore\cos\theta=\frac{3\times2+2\times(-12)+(-6)\times(-3)}{\sqrt{3^2+2^2+(-6)^2}\sqrt{2^2+(-12)^2+(-3)^2}}$
$=\frac{6-24+18}{\sqrt{49}\sqrt{157}}$
$=0$
$\Rightarrow\theta=\frac{\pi}{2}$
Thus, the angle between the given lines is $\frac{\pi}{2}.$
View full question & answer→Question 353 Marks
It the lines $\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{2\lambda}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}-6}{-5}$ are perpendicular, find the value of $\lambda.$
AnswerThe diraction of ratios of the lines, $\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{2\lambda}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}-6}{-5},$ are -3, 2k, 2 and 3k, 1, -5 respectiveiy.
It is know that two lines with direction ratios, $a_1, b_1, c_1$ and $a_2, b_2, c_2$, are perpendicular, if $a_1a_2+ b_1b_2+ c_1c_2= 0$
$\therefore -3 (3k) + 2k \times 1 + 2 (-5) = 0$
$\Rightarrow -9k + 2k - 10 = 0$
$\Rightarrow 7k = - 10$
$\Rightarrow\text{k}=\frac{-10}{7}$
Therefore, for $\text{k}=-\frac{10}{7},$ the given lines are perpendicular to each other.
View full question & answer→Question 363 Marks
Show that the line joining the origin to the point $(2, 1, 1)$ is perpendicular to the line determined by the points $(3, 5, -1)$ and $(4, 3, -1).$
AnswerHere,
A(0, 0, 0) and B(2, 1, 1)
C(3, 5, -1) and D(4, 3, -1)
Direction ratios of line $AB$
$a_1 = 2, b_1 = 1, c_1= 1$
Direction ratios of line $CD$
$a_2 = 2, b_2 = -2, c_2= 0$
Now,
$a_1a_2+ b_1b_2+ c_1c_2$
$= (2)(1) + (1)(-2) + (1)(0)$
$= 2 - 2 + 0$
$= 0$
Since, $a_1a_2+ b_1b_2+ c_1c_2= 0$, lines are perpendicular.
View full question & answer→Question 373 Marks
Find the vector and cartesian equations of a plane passing through the point (1, -1, 1) and normal to the line joining the points (1, 2, 5) and (-1, 3, 1)
AnswerSince the given plane passes through the point (1, -1, 1) and is normal to the line joining A(1, 2, 5) and B(-1, 3, 1)
$\overrightarrow{\text{n}}=\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(-\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}})-(\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}})$
$=-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$
We know that the vector equation of the passing through a point $\overrightarrow{\text{a}}$ and normal to $\overrightarrow{\text{n}}$ is,
$\overrightarrow{\text{r}}\cdot\overrightarrow{\text{n}}=\overrightarrow{\text{a}}\cdot\overrightarrow{\text{n}}$
Substituting $\overrightarrow{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\overrightarrow{\text{n}}=-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}},$ we get
$\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})$
$\Rightarrow\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=-2-1-4$
$\Rightarrow\overrightarrow{\text{r}}\cdot\big[-(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})\big]=-7$
$\Rightarrow\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=7$
For cartesian form, we need to substitute $\overrightarrow{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation.
Then, we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})=7$
$\Rightarrow2\text{x}-\text{y}+4\text{z}=7$
View full question & answer→Question 383 Marks
Write the equation of a plane which is at a distance of $5\sqrt{3}$ units from origin and the normal to which is equally inclined to coordinate axes.
AnswerLet $\alpha,\beta$ and $\gamma$ be the angles made by $\vec{\text{n}}$ with x, y and z-axes, respectively.
It is given that
$\alpha=\beta=\gamma$
$\Rightarrow\cos\alpha=\cos\beta=\cos\gamma$
$\Rightarrow\text{l}=\text{m}=\text{n},$ where l, m, n are direction cosines of $\vec{\text{n}}$.
But $\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\text{l}^2+\text{l}^2+\text{l}^2=1$
$\Rightarrow3\text{l}^2=1$
$\Rightarrow\text{l}^2=\frac{1}{3}$
$\Rightarrow\text{l}^2=\frac{1}{\sqrt{3}}$
So, $\text{l}=\text{m}=\text{n}=\frac{1}{\sqrt{3}}$
It is given that the length of the perpendicular of the plane from the origin, $\text{p}=5\sqrt{3}$
The normal form of the plane is $\text{lx}+\text{my}+\text{nz}=\text{p}$
$\Rightarrow\frac{1}{\sqrt{3}}\text{x}+\frac{1}{\sqrt{3}}\text{y}+\frac{1}{\sqrt{3}}\text{z}=5\sqrt{3}$
$\Rightarrow\text{x}+\text{y}+\text{z}=5\sqrt{3}\ (\sqrt{3})$
$\Rightarrow\text{x}+\text{y}+\text{z}=15.$
View full question & answer→Question 393 Marks
Find the distance of the point P(-1, -5, -10) from the point of intersection of the line joining the points A(2, -1, 2) and B(5, 3, 4) with the plane x - y + z = 5.
AnswerEquation of the line through the points A(2, -1, 2) and B(5, 3, 4) is $\frac{\text{x}-2}{5-2}=\frac{\text{y}+1}{3+1}=\frac{\text{z}-2}{4-2}=\text{r}$
$\Rightarrow\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-2}{2}=\text{r}$
$\Rightarrow\text{x}=3\text{r}+2,\text{ y}=4\text{r}-1,\text{ z}=2\text{r}+2$
Substituting these in the plane equation we get
$(3\text{r}+2)-(4\text{r}-1)+(2\text{r}+2)=5$
$\Rightarrow\text{r}=0$
$\Rightarrow\text{x}=2,\text{ y}=-1,\text{ z}=2$
Distance of (2, -1, 2) from (-1, -5, -10) is
$=\sqrt{(2-(-1))^2+(-1-(-5))^2+(2-(-10))^2}=\sqrt{3^2+4^2+12^2}$
$=\sqrt{169}=13$
View full question & answer→Question 403 Marks
Write the vector equation of the line passing through the point (1, -2, -3) and normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=5.$
AnswerThe required line is normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=5$ and it is parallel to the normal vector of the plane.
So, the required line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
It is given that the line passes through the point (1, -2, -3) whose position vector is given by $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
We know that the equation of the line passing through the point whose position vector is $\vec{\text{a}}$ and parrallel to the vector $\vec{\text{b}}$ is given by
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\vec{\text{r}}=\big(\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$
View full question & answer→Question 413 Marks
Write the distance between the parallel planes $2x − y + 3z = 4$ and $2x − y + 3z = 18$.
AnswerThe given equation are
$2x − y + 3z = 4$ ....(1)
The second equation of the plane is
$2x − y + 3z = 18$ .....(2)
We know that distance between two planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is $\frac{|\text{d}_2-\text{d}_1|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance is
$\frac{|18-4|}{\sqrt{2^2+(-1)^2+3^2}}$
$=\frac{|14|}{\sqrt{4+1+9}}$
$=\frac{14}{\sqrt{14}}$
$=\sqrt{14}\text{ units}$
View full question & answer→Question 423 Marks
The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle $\alpha.$ Prove that the equation of the plane in its new position is $\text{ax}+\text{by}\pm\bigg(\sqrt{\text{a}^2+\text{b}^2}\tan\alpha\bigg)\text{z}=0.$
AnswerEquation of the plane is ax + by = 0 .....(i)
$\therefore$ Equation of the plane after new position is
$\frac{\text{ax}\cos\alpha}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{\text{by}\cos\alpha}{\sqrt{\text{b}^2+\text{a}^2}}\pm\text{z}\sin\alpha=0$
Dividing both sides of above equation by cos $\alpha,$ we get
$\Rightarrow\frac{\text{ax}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{\text{by}}{\sqrt{\text{b}^2+\text{c}^2}}+\text{z}\tan\alpha=0$
$\Rightarrow\text{ax}+\text{by}\pm\text{z}\tan\alpha\sqrt{\text{a}^2+\text{b}^2}=0$ $\big($On multiplying with $\sqrt{\text{a}^2+\text{b}^2}\big)$
Hence proved.
View full question & answer→Question 433 Marks
Find the equation of the line passing through the points (1, 2, -4) and parallel to the line $\frac{\text{x}-3}{4}=\frac{\text{y}-5}{2}=\frac{\text{z}+1}{3}.$
AnswerThe direction ratios of the line parallel to line $\frac{\text{x}-3}{4}=\frac{\text{y}-5}{2}=\frac{\text{z}+1}{3}$ are proportional to 4, 2, 3.
Equation of the required line passing through the point (1, 2, -4) having direction ratios proportional to 4, 2, 3 is
$\frac{\text{x}-1}{4}=\frac{\text{y}-2}{2}=\frac{\text{z}-(-4)}{3}$
$=\frac{\text{x}-1}{4}=\frac{\text{y}-2}{2}=\frac{\text{z}+4}{3}$
View full question & answer→Question 443 Marks
Find the equation of a plane which meets the axes at A, B and C, given that the centroid of the triangle ABC is the point $(\alpha,\beta,\gamma)$
AnswerLet a, b and c be the intercepts of the given plane on the coordinate axes.
Then the plane meets the coordinate axes at
A(a, 0, 0), B(0, b, c) and C(0, 0, c)
Given that the centroid of the triangle $=(\alpha,\beta,\gamma)$
$\Rightarrow\Big(\frac{\text{a}+0+0}{3},\frac{0+\text{b}+0}{3},\frac{0+0+\text{c}}{3}\Big)=(\alpha,\beta,\gamma)$
$\Rightarrow\Big(\frac{\text{a}}{3},\frac{\text{b}}{3},\frac{\text{c}}{3}\Big)=(\alpha,\beta,\gamma)$
$\Rightarrow\frac{\text{a}}{3}=\alpha,\frac{\text{b}}{3}=\beta,\frac{\text{c}}{3}=\gamma$
$\Rightarrow\text{a}=3\alpha,\text{b}=3\beta,\text{c}=3\gamma\ ...(\text{i})$
The equation of the plane whose intercepts on the coordinate axes are a, b and c are
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
$\Rightarrow\frac{\text{x}}{3\alpha}+\frac{\text{y}}{3\beta}+\frac{\text{z}}{3\gamma}=1$ [From (i)]
$\Rightarrow\frac{\text{x}}{\alpha}+\frac{\text{y}}{\beta}+\frac{\text{z}}{\gamma}=3$
View full question & answer→Question 453 Marks
If the line drawn from the point (-2, -1, -3) meets a plane at right angle at the point (1, -3, 3), find the equation of the plane.
AnswerSince, the line drawn from the point (-2, -1, -3) meets a plane at right angle at the point (1, -3, 3) so, the plane passes through the point (1, -3, 3).
Also normal to plane is $(-3\hat{\text{i}}+2{\hat{\text{j}}}-6\hat{\text{k}})$
$\Rightarrow\vec{\text{a}}=\hat{\text{i}}-3{\hat{\text{j}}}+3\hat{\text{k}}$
And $\vec{\text{N}}=-3\hat{\text{i}}+2{\hat{\text{j}}}-6\hat{\text{k}}$
So, the equation of required plane is $(\vec{\text{r}}-\vec{\text{a}})\cdot\vec{\text{N}}=0$
$\Rightarrow\Big[(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})-(\hat{\text{i}}-3\hat{\text{j}}+3\hat{\text{k}})\Big]\cdot(3\hat{\text{i}}+2\hat{\text{j}}-6\hat{\text{k}})=0$
$\Rightarrow\Big[(\text{x}-1)\hat{\text{i}}+(\text{y}+3)\hat{\text{j}}+(\text{z}-3)\hat{\text{k}}\Big]\cdot(3\hat{\text{i}}+2\hat{\text{j}}-6\hat{\text{k}})=0$
$\Rightarrow-3\text{x}+3+2\text{y}+6-6\text{z}+18=0$
$\Rightarrow-3\text{x}+2\text{y}-6\text{z}=-27$
$\Rightarrow-3\text{x}+2\text{y}-6\text{z}-27=0$
View full question & answer→Question 463 Marks
Find the distance of the point (2, 4, -1) from the line $\frac{\text{x}+5}{1}=\frac{\text{y}+3}{4}=\frac{\text{z}-6}{-9}.$
AnswerLet P = (2, 4, -1)
In order to find the distance we need to find a point Q on the line.
So, let take this point as required point.
Also line is parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+4\hat{\text{j}}-9\hat{\text{k}}.$
Now, $\overrightarrow{\text{PQ}}=\big(-5\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}\big)-\big(2\hat{\text{i}}+4\hat{\text{j}}-\hat{\text{k}}\big)=-7\hat{\text{i}}-7\hat{\text{j}}+7\hat{\text{k}}$
$\vec{\text{b}}\times\overrightarrow{\text{PQ}}=\begin{bmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&4&-9\\-7&-7&7 \end{bmatrix}=-35\hat{\text{i}}+56\hat{\text{j}}+21\hat{\text{k}}$
$\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|=\sqrt{1225+3136+441}=\sqrt{4802}$
$\big|\vec{\text{b}}\big|=\sqrt{1+16+81}=\sqrt{98}$
$\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}=\frac{\sqrt{4802}}{\sqrt{98}}=7$
View full question & answer→Question 473 Marks
Find the angle between the plane:
2x - y + z = 4 and x + y + 2z = 3
AnswerWe know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between 2x - y + z = 4 and x + y + 2z = 3 is given by
$\cos\theta=\frac{(2)(1)+(-1)(1)+(1)(2)}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+1^2}}$
$=\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}}=\frac{3}{\sqrt{6}\sqrt{6}}$
$=\frac{3}{6}=\frac{1}{2}$
$\theta=\cos^{-1}\Big(\frac{1}{2}\Big)=\frac{\pi}{3}$
View full question & answer→Question 483 Marks
Show that the lines $\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ are perpendicular to each
Answerwe have
$\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$
These equations can be-written as
$\frac{\text{x}-5}{7}=\frac{\text{y}-(-2)}{-5}=\frac{\text{z}-0}{1}\dots(1)$
$\frac{\text{x}-0}{1}=\frac{\text{y}-0}{2}=\frac{\text{z}-0}{3}\dots(2)$
$\therefore\vec{\text{m}_1}=$ vector parallel to line (1) $=7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{m}}_2=$ vector parallel to line (2) $=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Now,
$\vec{\text{m}}_1\vec{\text{m}}_2=\big(7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=7-10+3$
$=0$
Hence, the given two lines are perpendicular to each other.
View full question & answer→Question 493 Marks
Find the vector and the cartesian equations of the line that passes through the points (3, -2, -5), (3, -2, 6).
AnswerLet $\vec{\text{a}}\ \text{and}\ \vec{\text{b}}$ be the position vectors of the points A(3, -2, -5) and B(3, -2, 6) respectively.
$\therefore\ \ \vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}\ \text{and}\ \vec{\text{b}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}$
$\therefore$ A vector along the line $=\overrightarrow{\text{AB}}=$ Position vector of point B - Position vector of point A
$\Rightarrow\ \ \overline{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}-3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}$
$\because$ Vector equation of the line is $\Big(\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\Big)$
$\therefore\ \ \vec{\text{r}}=3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}+\lambda\Big(11\hat{\text{k}}\Big)$
And another vector equation for the same line is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{AB}}= \vec{\text{r}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}+\lambda\Big(11\hat{\text{k}}\Big)$
Cartesian equation
Direction ratios of line AB are 3 - 3, -2 + 2, 6 + 5 = 0, 0, 11
$\therefore$ Equation of the line is $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}=\frac{\text{x}-3}{0}=\frac{\text{}\text{y}+2}{0}=\frac{\text{z}+5}{11}$
View full question & answer→Question 503 Marks
The cartesian equation of a line AB are $\frac{2\text{x}-1}{\sqrt{3}}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{3}.$ Find the direction cosines of a line parallel to AB.
AnswerWe have
$\frac{2\text{x}-1}{\sqrt{3}}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{3}$
The equation of the line AB can be re-written as
$\frac{\text{x}-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{3}$
$=\frac{\text{x}-\frac{1}{2}}{\sqrt{3}}=\frac{\text{y}+2}{4}=\frac{\text{z}-3}{6}$
Thus, the direction ratios of the line parallel to AB are proportional to 3, 4, 6.
Hence, the direction cosines of the line parallel to AB are proportional to
$\frac{\sqrt{3}}{\sqrt{(\sqrt{3})^2+4^2+6^2}},\frac{4}{\sqrt{(\sqrt{3})^2+4^2+6^2}},\frac{6}{\sqrt{(\sqrt{3})^2+4^2+6^2}}$
$=\frac{\sqrt{3}}{\sqrt{55}},\frac{4}{\sqrt{55}},\frac{6}{\sqrt{55}}$
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