Question 12 Marks
Write the equation of the plane parallel to XOY- plane and passing through the point (2, -3, 5).
AnswerThe equation of the plane parallel to the plane XOY is z = b .....(i), where b is a constant. It is given that this palne passes through (2, -3, 5).
So, 5 = b
Substituting this value in (i), we get z = 5, which is the required equation of the plane.
View full question & answer→Question 22 Marks
Find the vector equation one of following plane.
x + y - z = 5
AnswerGiven, equation of plane is,
x + y - z = 5
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})=5$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=5$
So,
Vector equation of the plane is $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=5$
View full question & answer→Question 32 Marks
Find the vector equation one of following plane.
2x - y + 2z = 8
AnswerGiven, equation of plane is,
2x - y + 2z = 8
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=8$
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=8$
So,
Vector equation of the plane is $\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=8$
View full question & answer→Question 42 Marks
Find the cartesian form of the equation of a plane whose vector equation is:
$\vec{\text{r}}\cdot\big(12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+5=0$
AnswerGiven the vector equation of a plane,
$\vec{\text{r}}\cdot\big(12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+5=0$
Let, $\vec{\text{r}}=\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)$
$\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)\big(12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+5=0$
$(\text{x})(12)+(\text{y})(-3)+(\text{z})(4)=0$
$12\text{x}-3\text{y}+4\text{z}+5=0$
Cartesian form of the equation of the plane of the plane is given by
$12\text{x}-3\text{y}+4\text{z}+5=0$
View full question & answer→Question 52 Marks
Find the equation of the plane passing through the following point:
(2, 3, 4), (-3, 5, 1) and (4, -1, 2)
AnswerThe equation of the plane passing through points (2, 3, 4), (-3, 5, 1) and (4, -1, 2) is given by,
$\begin{vmatrix}\text{x}-2&\text{y}-3&\text{z}-4\\-3-2&5-3&1-4\\4-2&-1-3&2-4\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-3&\text{z}-4\\-5&2&-3\\2&-4&-2\end{vmatrix}=0$
$\Rightarrow-16(\text{x}-2)-16(\text{y}-3)+16(\text{z}-4)=0$
$\Rightarrow(\text{x}-2)+(\text{y}-3)-(\text{z}-4)=0$
$\Rightarrow\text{x}+\text{y}-\text{z}=1$
View full question & answer→Question 62 Marks
If the equations of a line AB are $\frac{3-\text{x}}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-5}{4},$ write the direction ratios of a line parallel to AB.
AnswerWe have $\frac{3-\text{x}}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-5}{4}$ The equation of the line AB can be re-written as $\frac{\text{x}-3}{-1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-5}{4}$ Thus, the direction ratios of the line parllel to Ab are proportional to -1, -2, 4.
View full question & answer→Question 72 Marks
Write the cartesian and vector equations of x-axis.
AnswerSince x-axis passes through the point (0, 0, 0) having position vector $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ having direction ratios proportional to 1, 0, 0, the cartesian equation of x-axis is
$\frac{\text{x}-0}{1}=\frac{\text{y}-0}{0}=\frac{\text{z}-0}{0}$
$=\frac{\text{x}}{1}=\frac{\text{y}}{0}=\frac{\text{z}}{0}$
Also, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}+\lambda\big(\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)$
$=\lambda\hat{\text{i}}$
View full question & answer→Question 82 Marks
What are the direction cosines of Z-axis?
AnswerThe z-axis makes angles 90°, 90° and 0° with x, y and z axes, respectively.
Therefore, the direction cosines of x-axis are cos 90°, cos 90°, cos 0°, i.e. 0, 0, 1.
View full question & answer→Question 92 Marks
Write the value of k for which the line $\frac{\text{x}-1}{2}=\frac{\text{y}-1}{3}=\frac{\text{z}-1}{\text{k}}$ is perpendicular to the normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})=4.$
AnswerDirection ratios of the given line $\frac{\text{x}-1}{2}=\frac{\text{y}-1}{3}=\frac{\text{z}-1}{\text{k}}$ are proportional to 2, 3, k.
Direction ratios of the normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})=4$ are 2, 3, 4.
Given that these two are perpendicular.
⇒ (2) (2) + (3) (3) + (k) (4) (Because $a_1a_2 + b_1b_2 + c_1c_2 = 0$)
⇒ 4 + 9 + 4k = 0
⇒ 13 + 4k = 0
$\Rightarrow\text{k}=\frac{-13}{4}$.
View full question & answer→Question 102 Marks
Find the equation of a line parallel to x-axis and passing through the origin.
AnswerVector equation of a line is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
The direction cosines of the x-axis are (1, 0, 0). Equation of a line parallel to the x-axis and passing through the origin is
$\vec{\text{r}}=\big(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)+\lambda\big(1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)$
$\vec{\text{r}}=\lambda\hat{\text{i}}$
View full question & answer→Question 112 Marks
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}.$
AnswerThe normal vector is, $\vec{\text{n}}=3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}$
$\therefore\ \ \hat{\text{n}}=\frac{\vec{\text{n}}}{|\vec{\text{n}}|}=\frac{3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}}{\sqrt{(3)^2+(5)^2+(6)^2}}=\frac{3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}}{\sqrt{70}}$
It is known that the equation of the plane with position vector $\vec{\text{r}}$ is given by,
$\vec{\text{r}}.\hat{\text{n}}=\text{d}$
$\Rightarrow\ \vec{\text{r}}.\Bigg(\frac{3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}}{\sqrt{70}}\Bigg)=7$
This is the vector equation of the required plane.
View full question & answer→Question 122 Marks
If a line has direction ratios proportional to 2, -1, -2, then what are its direction consines?
AnswerIf a line has direction ratios proportional to 2, -1 and -2, then its direction cosines are
$\frac{2}{\sqrt{(2)^2+(-1)+(-2)^2}},\frac{-1}{\sqrt{(2)^2+(-1)+(-2)^2}},\frac{-2}{\sqrt{(2)^2+(-1)+(-2)^2}}$
$=\frac{2}{3},-\frac{1}{3},-\frac{2}{3}$
Thus, the direction cosines are $=\frac{2}{3},-\frac{1}{3},-\frac{2}{3}$.
View full question & answer→Question 132 Marks
Find the equation of the plane passing through the following point:
(1, 1, 1), (1, -1, 2) and (-2, -2, 2)
AnswerThe equation of the plane passing through points (1, 1, 1), (1, -1, 2) and (-2, -2, 2) is given by,
$\begin{vmatrix}\text{x}-1&\text{y}-1&\text{z}-1\\1-1&-1-1&2-1\\-2-1&-2-1&2-1\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-1&\text{y}-1&\text{z}-1\\0&-2&1\\-3&-3&1\end{vmatrix}=0$
$\Rightarrow1(\text{x}-1)-3(\text{y}-1)-16(\text{z}-1)=0$
$\Rightarrow\text{x}-1-3\text{y}+3-6\text{z}+6=0$
$\Rightarrow\text{x}-3\text{y}-6\text{z}+8=0$
View full question & answer→Question 142 Marks
Find the vector equation of the plane which is at a distance of 5 units from the orgin and its normal vector is $2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}.$
AnswerGiven:Normal vector, $\hat{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
Perpendicular distance, d = 5 units
The vector equation of a plane that is at a distance of 5 units from the origin and has its normal vector $\hat{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$ is as follows:
$\vec{\text{r}}.\hat{\text{n}}=\text{d}$
$\vec{\text{r}}.\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)=5.$
View full question & answer→Question 152 Marks
Write the distance of the point P(x, y, z) from XOY plane.
AnswerThe distance of point P(x, y, z) from the XOY plane is z.
View full question & answer→Question 162 Marks
If a line makes angles 90º, 135º, 45º with the x, y and z-axes respectively, find its direction cosines.
AnswerLet l, m, n be the direction cosines of the line with direction angles 90º, 135º, 45º.
$\therefore\text{l}=\cos90^\circ,\ \text{m}=\cos135^\circ=\cos(180^\circ-45^\circ)=-\cos45^\circ=-\frac{1}{\sqrt{2}},$
$\text{n}=\cos45^\circ=\frac{1}{\sqrt{2}}$
$\therefore$ direction cosines are $0,-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}.$
View full question & answer→Question 172 Marks
Determine the direction cosines of the normal to the plane and the distance from the origin.
2x + 3y - z = 5
Answer2x + 3y - z = 5 ...(1)
The direction ratios of normal are 2, 3, and -1.
$\therefore\ \sqrt{(2)^2+(3)^2+(-1)^2}=\sqrt{14}$
Dividing both sides of equation (1) by $\sqrt{14},$ we obtain
$\frac{2}{\sqrt{14}}\text{x}+\frac{3}{\sqrt{14}}\text{y}-\frac{1}{\sqrt{14}}\text{z}=\frac{5}{\sqrt{14}}$
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal to the plane are $\frac{2}{\sqrt{14}},\ \frac{3}{\sqrt{14}},\ \text{and}\ \frac{-1}{\sqrt{14}}$ and the distance of normal from the origin is $\frac{5}{\sqrt{14}}$ units.
View full question & answer→Question 182 Marks
Find the length of the perpendicular drawn from the origin to the plane $2x − 3y + 6z + 21 = 0$.
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
$\frac{|\text{ac}_1+\text{by}_1+\text{cz}_1+\text{d}|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|2(0)-3(0)+6(0)+21|}{\sqrt{2^2+(-3)^2+6^2}}$
$=\frac{|21|}{\sqrt{4+9+36}}$
$=\frac{21}{7}$
$=3\text{ units}$
View full question & answer→Question 192 Marks
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line $\text{5x – 25 = 14 – 7y = 35z.}$
AnswerEquation of given line is $\frac{\text{x - 5}}{1/5} = \frac{\text{y - 2}}{-1/7} = \frac{\text{z}}{1/35}$
Its DR's $\bigg\langle\frac{1}{5}, -\frac{1}{7}, \frac{1}{35}\bigg\rangle \text{ or } \langle7, -5, 1\rangle$
Equation of required line is
$\vec{\text{r}} = (\hat{\text{i}} + 2\hat{\text{j}} - \hat{\text{k}}) + \lambda (7\hat{\text{i}} - 5\hat{\text{j}} + \hat{\text{k}})$
View full question & answer→Question 202 Marks
Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.
AnswerFor x-intercepts, put y = 0 and z = 0 in the given eqution.
Then, we get
2x + 0 - 0 = 5
⇒ 2x = 5
$\Rightarrow\text{x}=\frac{5}{2}$
View full question & answer→Question 212 Marks
Write the angle between the line $\frac{\text{x}-1}{2}=\frac{\text{y}-2}{1}=\frac{\text{z}+3}{-\text{2}}$ and the plane x + y + 4 = 0.
AnswerThe given line is parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$ and the given plane is normal to the vector $\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$.
We know that the angle $\theta$ between the line and the plane is given by
$\sin\theta=\frac{\vec{\text{b}}.\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
$=\frac{\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)\big(\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}\big)}{|\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}||\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}|}$
$=\frac{1+2+0}{\sqrt{1+4+4}\sqrt{1+1+0}}$
$=\frac{3}{3\sqrt{2}}$
$=\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\sin^{-1}\Big(\frac{1}{\sqrt{2}}\Big)=45^\circ$
View full question & answer→Question 222 Marks
Determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0
Answer5y + 8 = 0
0x - 5y + 0z = 8 ....(1)
The direction ratios of normal are 0, -5, and 0.
$\therefore\ \sqrt{0+(-5)+0}=5$
Dividing both sides of equation (1) by 5, we obtain
$-\text{y}=\frac{8}{5}$
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal to the plane are 0, -1, and 0 and the distance of normal from the origin is $\frac{8}{5}$ units.
View full question & answer→Question 232 Marks
Write the inclination a line with z-axis, if its direction ratios are proportional to 0, 1, -1.
AnswerWe know that if a line has direction ratio (a, b, c), then the cosine of its angle with the z-axis is given by
$\cos\gamma=\frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
Suppose the inclination of the line with direction ratio (0, 1, -1) with z-axis is $\gamma$.
Now,
$\cos\gamma=\frac{1}{\sqrt{0+1+1}}$
$=-\frac{1}{\sqrt{2}}$
Implies that $\lambda=\frac{3\pi}{4}$
Hence, the inclination of the line with z-axis is $\frac{3\pi}{4}$.
View full question & answer→Question 242 Marks
Find the Cartesian equation of the following plane:$\vec{\text{r}}.\Big(2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}\Big)=1$
Answer$\vec{\text{r}}.\Big(2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}\Big)=1\ \ \ ....(1)$
For any arbitrary point P(x, y, z) on the plane, position vector $\vec{\text{r}}$ is given by,
$\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
Substituting the value of $\vec{\text{r}}$ in equation(1), we obtain
$\Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}-\text{z}\hat{\text{k}}\Big).\Big(2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}\Big)=1$
⇒ 2x + 3y - 4z = 1
This is the Cartesian equation of the plane.
View full question & answer→Question 252 Marks
Write the distance of the point (3, −5, 12) from X-axis?
AnswerThe distance of a general point (x, y, z) from x-axis is $\sqrt{\text{y}^2+\text{z}^2}$.
$\therefore$ Distance of the point (3, -5, 12) from x-axis $=\sqrt{(-5)^2+12^2}$
$=\sqrt{169}$
= $13\text{ units}$.
View full question & answer→Question 262 Marks
The x-coordinate of a point on the line joining the points P(2, 2, 1) and Q(5, 1, – 2) is 4. Find its z-coordinate.
AnswerEquation of line PQ is $\frac{\text{x - 2}}{3} = \frac{\text{y - 2}}{-1} = \frac{\text{z - 1}}{-3}$
Any point on the line is $(3\lambda + 2, -\lambda + 2, -3\lambda + 1)$
$3\lambda + 2 = 4 \Rightarrow \lambda = \frac{2}{3} \therefore \text{z coord.} = -3\bigg(\frac{2}{3}\bigg) + 1 = -1.$
Alternate Answer
Let R(4, y, z) lying on PQ divides PQ in the ratio k:1
$\Rightarrow 4 = \frac{\text{5k + 2}}{\text{k + 1}} \Rightarrow \text{k = 2}.$
$\therefore \text{z} = \frac{2(-2) + 1 (1)}{2 + 1} = \frac{-3}{3} = -1.$ View full question & answer→Question 272 Marks
What are the direction cosines of X-axis?
AnswerThe x-axis makes angles 0°, 90° and 90° with x, y and z axes, respectively.
Therefore, the direction cosines of x-axis are cos 90°, cos 0°, cos 90°, i.e. 1, 0, 0.
View full question & answer→Question 282 Marks
The equations of a line are given by $\frac{4-\text{x}}{3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}.$ Write the direction cosines of a line parallel to this line.
AnswerWe have$\frac{4-\text{x}}{3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}$
The equation of the given line can be re-written as.
$\frac{\text{x}-4}{-3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}$
The direction ratios of the line parallel to the given line are proportional to -3, 3, 6.
Hence, the direction cosines of the line parallel to the given line are proportional to
$\frac{-3}{\sqrt{(-3)^2+3^2+6^2}},\frac{-3}{\sqrt{(3)^2+3^2+6^2}},\frac{6}{\sqrt{(-3)^2+3^2+6^2}}$
$=\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}$
View full question & answer→Question 292 Marks
The x-coordinate of a point on the line joining the points P(2, 2, 1) and Q(5, 1, – 2) is 4. Find its z-coordinate.
AnswerEquation of line PQ is $\frac{\text{x - 2}}{3} = \frac{\text{y - 2}}{-1} = \frac{\text{z - 1}}{-3}$
Any point on the line is $(3\lambda + 2, -\lambda + 2, -3\lambda + 1)$
$3\lambda + 2 = 4 \Rightarrow \lambda = \frac{2}{3} \therefore \text{z coord.} = -3\bigg(\frac{2}{3}\bigg) + 1 = -1.$
Alternate Answer
Let R(4, y, z) lying on PQ divides PQ in the ratio k:1
$\Rightarrow 4 = \frac{\text{5k + 2}}{\text{k + 1}} \Rightarrow \text{k = 2}.$
$\therefore \text{z} = \frac{2(-2) + 1 (1)}{2 + 1} = \frac{-3}{3} = -1.$ View full question & answer→Question 302 Marks
Determine the direction cosines of the normal to the plane and the distance from the origin.
x + y + z
Answerx + y + z = 1 ...(1)
The direction ratios of normal are 1, 1, and 1
$\therefore\ \sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
Dividing both sides of equation (1) by $\sqrt{3},$ we obtain
$\frac{1}{\sqrt{3}}\text{x}+\frac{1}{\sqrt{3}}\text{y}+\frac{1}{\sqrt{3}}\text{z}=\frac{1}{\sqrt{3}}\ \ .....(2)$
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal are $\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}},\ \text{and}\ \frac{1}{\sqrt{3}}$ and the distance of normal from the origin is $\frac{1}{\sqrt{3}}$ units.
View full question & answer→Question 312 Marks
Write the vector equation of a line passing through a point having position vector $\vec{\alpha}$ and parallel to vector $\vec{\beta}.$
AnswerThe vector equation of the line passing the point having position vector $\vec{\alpha}$ and parallel to vector $\vec{\beta}$is $\vec{\text{r}}=\vec{\alpha}+\lambda\vec{\beta}.$
View full question & answer→Question 322 Marks
Write direction cosines of a line parallel to z-axis.
AnswerA line parallel to z-axis, makes an angle of 90°, 90° and 0° with the x, y, and z axes, respectively.
Thus, the direction cosines are given by
$\text{l}=\cos90^\circ=0$
$\text{m}=\cos90^\circ=0$
$\text{n}=\cos0^\circ=1$
Therefore, direction cosines of a line parallel to the z-axis 0, 0, 1.
View full question & answer→Question 332 Marks
Find the distance of the point (2, 3, 4) from the x-axis.
AnswerA general point (x, y, z) is at a distance $\sqrt{\text{y}^2+\text{z}^2}$ of from the x-axis.
$\therefore$ Distance of the point (2, 3, 4) from x-axis
$=\sqrt{3^2+4^2}=\sqrt{25}$
$=5\text{ units}$.
View full question & answer→Question 342 Marks
Show that the line joining the origin to the point $(2, 1, 1)$ is perpendicular to the line determined by the points $(3, 5, -1), (4, 3, -1).$
AnswerWe know that direction ratios of the line joining the origin $(0, 0, 0)$ to the point are $x_2 - x_1, y_2 - y_1, z_2 - z_1 = 2 - 0, 1 - 0, 1 - 0 = 2, 1, 1 = a_1, b_1, c_1$
Similarly, direction ratios of the line joining the points $(3, 5, -1)$ and $(4, 3, -1)$ are
$x_2 - x_1, y_2 - y_1, z_2 - z_1 = 4 - 3, 3 - 5, -1 - (-1) = 1, -2, 0 = a_2, b_2, c_2$
For these two lines, $a_1a_2 + b_1b_2 +c_1c_2$
$= 2(1) + 1(-2) + 1(0) = 2 - 2 + 0 = 0$
Therefore, the two given lines are perpendicular to each other.
View full question & answer→Question 352 Marks
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point: $(-6, 0, 0)$
Plane: $2x - 3y + 6z - 2 = 0$
AnswerIt is known that the distance between a point $p(x_1, y_1, z_1)$, and a plane ax + By + Cz = D, is given by,
$\text{d}=\Bigg|\frac{\text{A}\text{x}_1+\text{B}\text{y}_1+\text{C}\text{z}_1-D}{\sqrt{\text{A}^2+\text{B}^2+\text{C}^2}}\Bigg|\ \ \ ....(1)$
The given point is (-6, 0, 0) and the plane is 2x - 3y + 6z - 2 = 0
$\therefore\ \ \Bigg|\frac{2(-6)-3\times0+6\times0-2}{\sqrt{(2)^2+(-3)^2}+(6)^2}\Bigg|=\Big|\frac{-14}{\sqrt{49}}\Big|=\frac{14}{7}=2.$
View full question & answer→Question 362 Marks
Write the formula for the shortest distance between the lines$\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}.$
AnswerThe shortest distance d between the parallel lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}$ is given by
$\text{d}=\frac{\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|}{\big|\vec{\text{b}}\big|}$
View full question & answer→Question 372 Marks
Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
AnswerThe equation of the plane ZOX is
y = 0
Any plane parallel to it is of the form, y = a
Since the y-intercept of the plane is 3,
$\therefore$ a = 3
Thus, the equation of the required plane is y = 3.
View full question & answer→Question 382 Marks
A line passes through the point with position vector $2\hat{\text{i}} - 3\hat{\text{j}} + 4\hat{\text{k}}$ and is perpendicular to the plane $\vec{\text{r}}. (3\hat{\text{i}} + 4\hat{\text{j}} - 5\hat{\text{k}}) = 7.$ Find the equation of the line in cartesian and vector forms.
AnswerVector form: $\vec{\text{r}} = (2\hat{\text{i}} - 3\hat{\text{j}} + 4\hat{\text{k}}) + \lambda (3\hat{\text{i}} + 4\hat{\text{j}} - 5\hat{\text{k}})$
Cartesian form: $\frac{\text{x - 2}}{3} = \frac{\text{y + 3}}{4} = \frac{\text{z - 4}}{-5}$
View full question & answer→Question 392 Marks
Define direction cosines of a direction line.
AnswerThe direction cosines of a direction line segment are the cosines of the direction angles of the line segment. Let two points $A(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ define the directed line segment AB.
The direction cosines of AB are given by cos
$\alpha=\frac{\text{x}_2-\text{x}_1}{\text{d}}$
$\cos\beta=\frac{\text{y}_2-\text{y}_1}{\text{d}}$
$\cos\gamma=\frac{\text{z}_2-\text{z}_1}{\text{d}}$
Here, d is the distance between A and B.
View full question & answer→Question 402 Marks
Show that the line through the points $(1, -1, 2), (3, 4, –2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6).$
AnswerWe know that direction ratios of the line joining the points $A(1, -1, 2)$ and $B(3, 4, -2)$ are
$x_2 - x_1, y_2 - y_1, z_2 - z_1$
$\Rightarrow 3 - 1, 4 - (-1), -2 - 2$
$\Rightarrow 2, 5, -4 = a_1, b_1, c_1$
Again, direction ratios of the line joining the points $C(0, 3, 2)$ and $D(3, 5, 6)$ are
$x_2 - x_1, y_2 - y_1, z_2 - z_1$
$\Rightarrow 3 - 0, 5 - 3, 6 - 2$
$\Rightarrow 3, 2, 4 = a_2, b_2, c_2 ($say$)$
For lines $AB$ and $CD, a_1a_2 + b_1b_2 + c_1c_2$
$= 2 \times 3 + 5 \times 2 + (-4) \times 4 = 6 + 10 - 16 = 0$
Therefore, line $AB$ is perpendicular to line $CD.$
View full question & answer→Question 412 Marks
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line $\text{5x – 25 = 14 – 7y = 35z.}$
AnswerEquation of given line is $\frac{\text{x - 5}}{1/5} = \frac{\text{y - 2}}{-1/7} = \frac{\text{z}}{1/35}$
Its DR's $\bigg\langle\frac{1}{5}, -\frac{1}{7}, \frac{1}{35}\bigg\rangle \text{ or } \langle7, -5, 1\rangle$
Equation of required line is
$\vec{\text{r}} = (\hat{\text{i}} + 2\hat{\text{j}} - \hat{\text{k}}) + \lambda (7\hat{\text{i}} - 5\hat{\text{j}} + \hat{\text{k}})$
View full question & answer→Question 422 Marks
Write the distance of the plane $\vec{\text{r}}.(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=12$ form the origin.
AnswerThe given equation of the plane is
$\vec{\text{r}}.(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=12$ or $\vec{\text{r}}.\vec{\text{n}}=-6$, where $\vec{\text{n}}=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{4+1+4}=3$
For reducing the given equation to normal form, we need to divide both sides by $|\vec{\text{n}}|$.
Then, we get $\vec{\text{r}}.\frac{\vec{\text{n}}}{|\vec{\text{n}}|}=\frac{12}{|\vec{\text{n}}|}$
$=\vec{\text{r}}.\Big(\frac{2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}}{3}\Big)=\frac{12}{3}$
$\Rightarrow\vec{\text{r}}.\Big(\frac{2}{3}\hat{\text{i}}+\frac{1}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}\Big)=4,\ ...(1)$
The eqution of the plane normal from is.
$\vec{\text{r}}.\text{n}=\text{d}\ ....(2)$
(where d is the distance of the plane from the origin)
Comparing (1) and (2),
Length of the perpendicular from the origin to the plane = d = 4 units.
View full question & answer→Question 432 Marks
Write the vector equation of a line given by $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}.$
AnswerWe have
$\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}$
The given line passes through the point (5, -4, 6) and has direction ratios proportional to 3, 7, 2.
Vector equation of the given line passing through the point having position vector $\vec{\text{a}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}$ and parallel to a vector $\vec{\text{b}}=3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}$ is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\vec{\text{r}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}+\lambda\big(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}\big)$
View full question & answer→Question 442 Marks
Write a vector normal to the plane $\vec{\text{r}}=\text{l}\vec{\text{b}}+\text{m}\vec{\text{c}}$ .
AnswerThe equation of the given plane is
$\vec{\text{r}}=\text{l}\vec{\text{b}}+\text{m}\vec{\text{c}}$
So, the plane passes parallel to the vectors $\vec{\text{b}}$ and $\vec{\text{c}}$.
So, the vector normal to the plane is $\vec{\text{b}}\times\vec{\text{c}}$.
View full question & answer→Question 452 Marks
The x-coordinate of a point on the line joining the points P(2, 2, 1) and Q(5, 1, – 2) is 4. Find its z-coordinate.
AnswerEquation of line PQ is $\frac{\text{x - 2}}{3} = \frac{\text{y - 2}}{-1} = \frac{\text{z - 1}}{-3}$
Any point on the line is $(3\lambda + 2, -\lambda + 2, -3\lambda + 1)$
$3\lambda + 2 = 4 \Rightarrow \lambda = \frac{2}{3} \therefore \text{z coord.} = -3\bigg(\frac{2}{3}\bigg) + 1 = -1.$
Alternate Answer
Let R(4, y, z) lying on PQ divides PQ in the ratio k:1
$\Rightarrow 4 = \frac{\text{5k + 2}}{\text{k + 1}} \Rightarrow \text{k = 2}.$
$\therefore \text{z} = \frac{2(-2) + 1 (1)}{2 + 1} = \frac{-3}{3} = -1.$ View full question & answer→Question 462 Marks
Write the cartesian and vector equations of z-axis.
AnswerSince z-axis passes through the point (0, 0, 0) having position vector $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=0\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}$ having direction ratios proportional to 0, 0, 1, the cartesian equation of z-axis is
$\frac{\text{x}-0}{0}=\frac{\text{y}-0}{0}=\frac{\text{z}-0}{1}$
$=\frac{\text{x}}{0}=\frac{\text{y}}{0}=\frac{\text{z}}{1}$
Also, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}+\lambda\big(0\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}\big)$
$=\lambda\hat{\text{k}}$
View full question & answer→Question 472 Marks
Write the coordinate axis to which the line $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-1}{0}$ is perpendicular.
AnswerWe have
$\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-1}{0}$
The given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}+0\hat{\text{k}}$
Let $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ be perpendicular to the given line.
Now,
$3\text{x}+4\text{y}+0\text{z}=0$
It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).
Hence, the given line is perpendicular to z-axis.
View full question & answer→Question 482 Marks
Write the position vector of the point where the line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ meets the plane $\vec{\text{r}}.\vec{\text{n}}=0$.
AnswerThe given equation of the plane is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\ ....(1)$
Given equation of the plane is
$\vec{\text{r}}.\vec{\text{n}}=0$
$\Rightarrow\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big)\vec{\text{n}}=0$ [ From (1) ]
$\Rightarrow\vec{\text{a}}.\vec{\text{n}}+\lambda\vec{\text{b}}.\vec{\text{n}}=0$
$\Rightarrow\lambda=-\Big(\frac{\vec{\text{a}}.\vec{\text{n}}}{\vec{\text{b}}.\vec{\text{n}}}\Big)$
Substituting this in (1), we get
$\vec{\text{r}}=\vec{\text{a}}-\Big(\frac{\vec{\text{a}}.\vec{\text{n}}}{\vec{\text{b}}.\vec{\text{n}}}\Big)\vec{\text{b}},$ which is the required position vector that lies both on the line and the plane.
View full question & answer→Question 492 Marks
Write the general equation of a plane parallel to X-axis.
AnswerThe general equation of a plane is
ax + by + cz + d = 0 .....(i)
This plane is parallel to the X-axis.
It means that this plane passes through the point (0, y, z). So,
a(0) + by + cz + d = 0
⇒ by + cz + d = 0
View full question & answer→Question 502 Marks
Find the Cartesian equation of the following plane:
$\vec{\text{r}}.\Big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\Big)=2$
AnswerIt is given that equation of the plane is
$\vec{\text{r}}.\Big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\Big)=2\ \ \ ....(1)$
For any arbitrary point P(x, y, z) on the plane, position vector $\vec{\text{r}}$ is given by $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}-\text{z}\hat{\text{k}}$
Substituting the value of $\vec{\text{r}}$ in equation (1), we obtain
$\Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\Big).\Big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\Big)=2$
⇒ x + y - z = 2
This is the Cartesian equation of the plane.
View full question & answer→