Question 512 Marks
Write the ratio in which the line segment joining (a, b, c) and (-a, -b, -c) is divided by the xy-plane.
AnswerSuppose the line segment joining the points (a, b, c) and (-a, -c, -b) is divided by the XY-plane at a point R in the ratio $\lambda:1$.
Coordinates of R are
$\Big(\frac{\lambda(-\text{a})+1(\text{a})}{\lambda+1},\lambda(-\text{c})+\frac{1(\text{b})}{\lambda+1},\frac{\lambda(-\text{b})+1(\text{c})}{\lambda+1}\Big)$
Since R lies on XY-plane, Z-coordinate of R must be zero.
$\Rightarrow\frac{\lambda(-\text{b})+1(\text{c})}{\lambda+1}=0=\frac{\text{c}}{\text{b}}$
Thus, the required ratio is c/b : 1 or c : b.
Hence, the XY-plane divided the line in the ratio c : b.
View full question & answer→Question 522 Marks
Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.
AnswerLet P(2, 3, 4), Q(-1, -2, 1), R(5, 8, 7) be given points.
The direction ratios of PQ are -1 - 2, -2 - 3, 1 - 4 i.e. -3, -5, -3
The direction ratios of PR are 5 - 2, 8 - 3, 7 - 4 i.e. 3, 5, 3
Since $\frac{-3}{3}=\frac{-5}{5}=\frac{-3}{3}$
$\therefore$ lines PQ and PR are parallel.
But P is a common point on both the lines points
$\therefore$ P, Q, R are collinear.
View full question & answer→Question 532 Marks
Find the vector equation of a plane passing throught a point with position $2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and perpendicular to the vector $4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
AnswerWe know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
Substituting $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{n}}=4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},$ we get
$\vec{\text{r}}\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)=\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{r}}\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)=8-2-3$
$\Rightarrow\vec{\text{r}}\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)=3$
View full question & answer→Question 542 Marks
If a line has direction ratios 2, -1, -2, determine its direction cosines.
AnswerLet the direction cosines of a line be l, m and n. Now,$\text{l}=\frac{2}{\sqrt{2^2+(-1)^2+(-2)^2}}=\frac{2}{3}$
$\text{m}=\frac{-1}{\sqrt{2^2+(-1)^2+(-2)^2}}=\frac{-1}{3}$ $\text{n}=\frac{-2}{\sqrt{2^2+(-1)^2+(-2)^2}}=\frac{-2}{3}$ $\therefore$ The direction consines of the line are $\frac{2}{3},\frac{-1}{3},\frac{-2}{3}.$
View full question & answer→Question 552 Marks
Write the distances of the point (7, -2, 3) from XY, YZ and XZ-planes.
AnswerThe distance of a general point P (x, y, z) from XY-plane is z.
Thus, distance of (7, -2, 3) from XY-plane is 3.
Similarly, the distance of P (x, y, z) from YZ-plane is x.
Thus, distance of (7, -2, 3) from YZ-plane is 7.
The distance of P (x, y, z) from XZ-plane is y.
Thus, distance of (7, -2, 3) from XZ-plane is 2.
View full question & answer→Question 562 Marks
Find the equation of the plane passing through the following point:
(0, -1, 0), (3, 3, 0) and (1, 1, 1)
AnswerThe equation of the plane passing through points (0, -1, 0), (3, 3, 0) and (1, 1, 1) is given by,
$\begin{vmatrix}\text{x}-0&\text{y}+1&\text{z}-0\\3-0&3+1&0-0\\1-0&1+1&1-0\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-0&\text{y}+1&\text{z}-0\\3&4&0\\1&2&1\end{vmatrix}=0$
$\Rightarrow4\text{x}-3(\text{y}+1)+2\text{z}=0$
$\Rightarrow4\text{x}-3\text{y}+2\text{z}=3$
View full question & answer→Question 572 Marks
Answer each of the following questions in one word or one sentence or as per exact requirement of the quetion:
Find the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane $\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2.$
AnswerThe required plane passes through $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ and is parallel to the plane $\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2.$
So, it is normal to the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ which is normal to the given plane.
Hence, the vector equation of the required plane is
$\big[\vec{\text{r}}-\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=0\big]\ \big[\big(\vec{\text{r}}-\vec{\text{a}}\big).\vec{\text{n}}=0\big]$
$\Rightarrow\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{a}+\text{b}+\text{c}$
Thus, the vector equation of the required plane is $\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{a}+\text{b}+\text{c}.$
View full question & answer→Question 582 Marks
If line makes angle $\alpha,\beta$ and $\gamma$ with the coordinate axes, find the value of $\cos2\alpha+\cos2\beta+\cos2\gamma$.
AnswerIt is given that the line makes angles $\alpha,\beta,\gamma$ with the coordinate axis.
$\therefore\text{l}=\cos\alpha,\text{m}=\cos\beta$ and $\text{n}=\cos\gamma$
$\Rightarrow\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\ ......(1)$
Now,
$\cos2\alpha+\cos2\beta+\cos2\gamma$
$=(2\cos^2\alpha-1)+(2\cos^2\beta-1)+(2\cos^2\gamma-1)$
$=2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)-3$
$=2(1)-3$
$=-1$
View full question & answer→Question 592 Marks
Find the cartesian form of the equation of a plane whose vector equation is:
$\vec{\text{r}}\cdot\big(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)=9$
AnswerHere, equation of the plane is,
$\vec{\text{r}}\cdot\big(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)=9$
Let $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}},$ then
$\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)\big(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)=9$
$(\text{x})(-1)+(\text{y})(1)+(\text{z})(2)=9$
$-\text{x}+\text{y}+2\text{z}=9$
Cartesian form of the equation of the plane is,
$-\text{x}+\text{y}+2\text{z}=9$
View full question & answer→Question 602 Marks
Find the value of k so that the lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular to each other.
AnswerTwo lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular.
$\frac{\text{x}}{1}=\frac{\text{y}}{-1}=\frac{\text{z}}{\frac{1}{\text{k}}} \ ...(1)$
$\frac{\text{x}-2}{1}=\frac{\text{y}+\frac{1}{2}}{\frac{1}{2}}=\frac{\text{z}-1}{-1} \ ...(2)$
On comparing with $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$
we get,
$\text{x}_1=0, \ \text{y}_1=0, \ \text{z}_1=0$
$\& \ \text{x}_2=2, \ \text{y}_2=\frac{-1}2, \ \text{z}=1$
$\text{a}_1=+1, \ \text{b}_1=-1, \ \text{c}_1=\frac{1}{\text{k}}$
$\& \ \text{a}_2=1, \ \text{b}_2=\frac{1}{2}, \ \text{c}_2=1 $
Since two lines are perpendicular, therefore
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$\Rightarrow1\times1=(-1)\times\frac{1}{2}+\frac{1}{\text{k}}\times(-1)=0$
$\Rightarrow1-\frac{1}{2}-\frac{1}{\text{k}}=0$
$\Rightarrow1-\frac{\text{k}-2}{2\text{k}}=0$
$\Rightarrow2\text{k}-\text{k}-2=0$
$\text{k}=2$
View full question & answer→Question 612 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 622 Marks
Find the equation of the line passing through the points (2, -1, 3) and parallel to the line $\vec{\text{r}}=\big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}\big).$
AnswerThe given line is parallel to the vector $2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$ and the required line is parallel to the given line.
So, the required line is parallel to the vector $2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$
hence, the equation of the required line passing through the points (2, -1, 3) and parallel to the vector
$2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$ is $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}\big)$
View full question & answer→Question 632 Marks
Write the ratio in which YZ-plane divides the segment joining P(-2, 5, 9) and Q(3, -2, 4).
AnswerLet the YZ-plane divides the line segment joining points P(-2, 5, 9) and Q(3, -2, 4) in the ratio k : 1.
Using the setion formula, the coordinates of the point of intersection are given by
$\Big(\frac{\text{k}(3)-2}{\text{k}+1},\frac{\text{k}(-2)+5}{\text{k}+1},\frac{\text{k}(4)+9}{\text{k}+1}\Big)$
On the YZ-plane, the X-coordinate of any point is zero.
$\frac{\text{k}(3)-2}{\text{k}+1}=0$
Implies that 3k - 2 = 0
Implies that $\text{k}=\frac{2}{3}$
Thus, the YZ-plane divides the line segment formed by joining the given points in the ratio 2 : 3 internally.
View full question & answer→Question 642 Marks
Find the Cartesian equation of the following plane:$\vec{\text{r}}.\Big[(\text{s - 2t})\hat{\text{i}}+(3-\text{t})\hat{\text{j}}+(2\text{s + t})\hat{\text{k}}\Big]=15$
Answer$\vec{\text{r}}.\Big[(\text{s - 2t})\hat{\text{i}}+(3-\text{t})\hat{\text{j}}+(2\text{s + t})\hat{\text{k}}\Big]=15$
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
$\Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\Big).\Big[(\text{s - 2t})\hat{\text{i}}+(3-\text{t})\hat{\text{j}}+(2\text{s + t})\hat{\text{k}}\Big]=15\ \ \ .....(1)$
⇒ (s - 2t)x + (3 - t)y + (2s + t)z = 15
This is the Cartesian equation of the given plane.
View full question & answer→Question 652 Marks
Answer the following quations in one word or one sentence or as per exact requirement of the question:
Write the distance of a point P(a, b, c) from x-axis.
AnswerWe know that a general point (x, y, z) has distance $\sqrt{\text{y}^2+\text{z}^2}$
$\therefore$ Distance of a point P(x, y, z) from x-axis $=\sqrt{\text{b}^2+\text{c}^2}$.
View full question & answer→Question 662 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 672 Marks
Find the vector equation of a plane which is at a distance of 3 units from the origin and has $\hat{\text{k}}$ as the unit vector normal to it.
AnswerHere, it is given that, the required plane is at a distance of 3 unit from origin and k is unit vector normal to it. we know that, vector equation of a plane normal to unit vector $\hat{\text{n}}$ and at distance d from origin, is
$\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$
So, here d = 3 units
$\hat{\text{n}}=\hat{\text{k}}$
The equation of the required plane is,
$\vec{\text{r}}\cdot\hat{\text{k}}=3$
View full question & answer→Question 682 Marks
Write the coordinates of the projection of point P(2, -3, 5) on Y-axis.
AnswerThe coordinates of the projection of the point P(2, -3, 5) on Y-axis are (0, -3, 0) as both x and z coordinates of each point on the y-axis are equal to zero.
View full question & answer→Question 692 Marks
Find the equations of the planes that passes through three points.
(1, 1, -1), (6, 4, -5), (-4, -2, 3).
AnswerThe given points are A(1, 1, -1), B(6, 4, -5), and C(-4, -2, 3).
$\begin{vmatrix}1&1&-1\\6&4&-5\\-4&-2&3\end{vmatrix}$
= 1(12 - 10) - 1(18 - 20) - 1(-12 + 16)
= 2 + 2 - 4 = 0
Since A, B, C are collinear points, there will be infinite number of
planes passing through the given points.
View full question & answer→Question 702 Marks
Find the intercepts cut off by the plane 2x + y - z = 5.
Answer2x + y - z = 5 ....(1)
Dividing both sides of equation (1) by 5, we obtain
$\frac{2}{5}\text{x}+\frac{\text{y}}{5}-\frac{\text{z}}{5}=1$
$\Rightarrow\frac{\text{x}}{\frac{5}{2}}\text{x}+\frac{\text{y}}{5}+\frac{\text{z}}{-5}=1\ \ ...(2)$
It is known that the equation of a plane in intercept form $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1,$
where a, b, c are the intercepts cut off by the plane at x, y, and z axes respectively.
Therefore, for the given equation,
$\text{a}=\frac{5}{2},\ \text{b}=5,\ \text{and}\ \text{c}=-5$
Thus, the intercepts cut off by the plane are $\frac{5}{2},\ 5,\ \text{and}\ -5.$
View full question & answer→Question 712 Marks
Write the cartesian and vector equations of y-axis.
AnswerSince y-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 $\vec{\text{b}}=0\hat{\text{i}}+1\hat{\text{j}}+0\hat{\text{k}}$ having direction ratios proportional to 0, 1, 0, the cartesian equation of y-axis is
$\frac{\text{x}-0}{1}=\frac{\text{y}-0}{0}=\frac{\text{z}-0}{0}$
$=\frac{\text{x}}{0}=\frac{\text{y}}{1}=\frac{\text{z}}{0}$
Also, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}+\lambda\big(0\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}\big)$
$=\lambda\hat{\text{j}}$
View full question & answer→Question 722 Marks
Write the value of $k$ for which the planes $x - 2y + kz = 4$ and $2x + 5y - z = 9$ are perpendicular.
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 prependicular to each of $a_1a_2 + b_1b_2 + c_1c_2 = 0$
The given planes are $x - 2y + kz = 4$ and $2x + 5y - z = 9$
$\Rightarrow a_1 = 1; b_1 = -2; c_1 = 5; a_2 = 2; b_2 = 5; c_2 = -1$
It is given that the given planes are perpendicular.
$\Rightarrow a_1a_2 + b_1b_2 + c_1c_2 = 0$
$\Rightarrow (1)(2) + (-2)(5) + (k)(-1) = 0$
$\Rightarrow 2 - 10 - k = 0$
$\Rightarrow -8 - k = 0$
$\Rightarrow k = -8$
View full question & answer→Question 732 Marks
Find the Cartesian equations of the line which passes through the point (-2, 4, -5) and is parallel to the line $\frac{\text{x}+3}{3}=\frac{4-\text{y}}{5}=\frac{\text{z}+8}{6}.$
AnswerThe equation of the given line is $\frac{\text{x}+3}{3}=\frac{4-\text{y}}{5}=\frac{\text{z}+8}{6}$It can be re-written as
$\frac{\text{x}+3}{3}=\frac{\text{y}-4}{-5}=\frac{\text{z}+8}{6}$
Since the required line is parallel to the given line, the direction ratios of the required line are proportional to 3, -5, 6.
Hence, the cartesian equations of the line passing through the point (-2, 4, -5) and parallel to a vector having direction ratios proportional to 3, -5, 6 is $\frac{\text{x}+2}{3}=\frac{\text{y}-4}{-5}=\frac{\text{z}+5}{6}.$
View full question & answer→Question 742 Marks
Find the value of k so that the lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular to each other.
AnswerTwo lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular.
$\frac{\text{x}}{1}=\frac{\text{y}}{-1}=\frac{\text{z}}{\frac{1}{\text{k}}} \ ...(1)$
$\frac{\text{x}-2}{1}=\frac{\text{y}+\frac{1}{2}}{\frac{1}{2}}=\frac{\text{z}-1}{-1} \ ...(2)$
On comparing with $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$
we get,
$\text{x}_1=0, \ \text{y}_1=0, \ \text{z}_1=0$
$\& \ \text{x}_2=2, \ \text{y}_2=\frac{-1}2, \ \text{z}=1$
$\text{a}_1=+1, \ \text{b}_1=-1, \ \text{c}_1=\frac{1}{\text{k}}$
$\& \ \text{a}_2=1, \ \text{b}_2=\frac{1}{2}, \ \text{c}_2=1 $
Since two lines are perpendicular, therefore
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$\Rightarrow1\times1=(-1)\times\frac{1}{2}+\frac{1}{\text{k}}\times(-1)=0$
$\Rightarrow1-\frac{1}{2}-\frac{1}{\text{k}}=0$
$\Rightarrow1-\frac{\text{k}-2}{2\text{k}}=0$
$\Rightarrow2\text{k}-\text{k}-2=0$
$\text{k}=2$
View full question & answer→Question 752 Marks
Write the equation of the plane corntaining the lines $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ and $\vec{\text{r}}=\vec{\text{a}}+\mu\vec{\text{c}}$.
AnswerThe given equation of the plane is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ and $\vec{\text{r}}=\vec{\text{a}}+\mu\vec{\text{c}}$.
So, the plane passes through the vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ and $\vec{\text{c}}$.
So, the plane passes through the vector $\vec{\text{a}}$ whose normal vector is $\vec{\text{b}}\times\vec{\text{a}}$
(It means that $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{a}}$)
So, the eqution of the plane in scalar product from is
$(\vec{\text{r}}-\vec{\text{a}}).\vec{\text{n}}=0$
$\Rightarrow(\vec{\text{r}}-\vec{\text{a}}).(\vec{\text{b}}\times\vec{\text{c}})=0$
View full question & answer→Question 762 Marks
Write the coordinates of the projection of point P(x, y, z) on XOZ-plane.
AnswerThe projection of the point P(x, y, z) on XOZ-plane is (x, 0, z) as Y-coordinates of any point on XOZ-plane are equal to zero.
View full question & answer→Question 772 Marks
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point: $(3, -2, 1)$
Plane: $2x - y + 2z + 3 = 0$
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 plane is (3, -2, 1) and the plane is 2x - y + 2z + 3 = 0
$\therefore\ \ \text{d}=\Bigg|\frac{2\times3+(-1)\times(-2)+2\times1+3}{\sqrt{(2)^2+(-1)^2+(2)^2}}\Bigg|=\Big|\frac{13}{3}\Big|=\frac{13}{3}$
View full question & answer→Question 782 Marks
Find the vector equation one of following plane.
x + y = 3
AnswerGiven, equation of plane is,
x + y = 3
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(\hat{\text{i}}+\hat{\text{j}})=3$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}})=3$
So,
Vector equation of the plane is $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}})=3$
View full question & answer→Question 792 Marks
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point: $(2, 3, -5)$
Plane: $x + 2y - 2z - 9 = 0$
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 plane is (2, 3, -5) and the plane is x + 2y - 2z = 9
$\therefore\ \ \text{d}=\Bigg|\frac{2+2\times3-2(-5)-9}{\sqrt{(1)^2+(2)^2+(-2)^2}}\Bigg|=\frac{9}{3}=3$
View full question & answer→Question 802 Marks
Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).
AnswerThe equation of the plane passing through (a, 0, 0), (0, b, 0) and (0, 0, c) is
$\begin{vmatrix}\text{x}-\text{a} & \text{y}-0&\text{z}-0 \\ 0-\text{a} & \text{b}-0 & 0 - 0 \\ 0-\text{a}&0-0&\text{C}-0 \end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-\text{a} & \text{y}&\text{z} \\ -\text{a} & \text{b} & 0 - 0 \\ -\text{a}& 0 &\text{C} \end{vmatrix}=0$
$\Rightarrow\ \text{bc}(\text{x}-\text{a})+\text{acy}+\text{abz}=0$
$\Rightarrow\ \text{bcx}+\text{acy}+\text{abz}=\text{abc}$
Dividing the equationg by abc, we get
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
View full question & answer→Question 812 Marks
Write the equation of the plane $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ in scalar product from.
AnswerThe given equation of the plane is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$
So, the plane passes through the vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ and $\vec{\text{c}}$.
So, the plane passes through the vector $\vec{\text{a}}$ whose normal vector is $\vec{\text{b}}\times\vec{\text{a}}$
(It means that $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{a}}$)
So, the eqution of the plane in scalar product from is
$(\vec{\text{r}}-\vec{\text{a}}).\vec{\text{n}}=0$
$\Rightarrow(\vec{\text{r}}-\vec{\text{a}}).(\vec{\text{b}}\times\vec{\text{c}})=0$
View full question & answer→Question 822 Marks
Find the vector equation of a plane which is at a distance of 5 unit from the origin and which is normal to the vector $\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
AnswerIt is given that the normal vector, $\vec{\text{n}}=\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
Now, $\text{n}=\frac{\vec{\text{n}}}{|\vec{\text{n}}|}$
$=\frac{\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}}{\sqrt{1+4+4}}$
$=\frac{\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}}{3}$
$=\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}$
The equation of a plane in normal form is
$\vec{\text{r}}\cdot{\text{n}}={\text{d}}$ (where d is the distance of the plane from the origin)
Substituting $\text{n}=\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}$ and d = 5
Here,
$\vec{\text{r}}\cdot\Big(\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}\Big)=5$
View full question & answer→Question 832 Marks
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point: $(0, 0, 0)$
Plane: $3x - 4y + 12z = 3$
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 (0, 0, 0) and the plane is 3x - 4y + 12z = 3
$\therefore\ \ \text{d}=\Bigg|\frac{3\times0-4\times0+12\times0-3}{\sqrt{(3)^2+(-4)^2+(12)^2}}\Bigg|=\frac{3}{\sqrt{169}}=\frac{3}{13}$
View full question & answer→Question 842 Marks
Find the value of k so that the lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular to each other.
AnswerTwo lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular.
$\frac{\text{x}}{1}=\frac{\text{y}}{-1}=\frac{\text{z}}{\frac{1}{\text{k}}} \ ...(1)$
$\frac{\text{x}-2}{1}=\frac{\text{y}+\frac{1}{2}}{\frac{1}{2}}=\frac{\text{z}-1}{-1} \ ...(2)$
On comparing with $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$
we get,
$\text{x}_1=0, \ \text{y}_1=0, \ \text{z}_1=0$
$\& \ \text{x}_2=2, \ \text{y}_2=\frac{-1}2, \ \text{z}=1$
$\text{a}_1=+1, \ \text{b}_1=-1, \ \text{c}_1=\frac{1}{\text{k}}$
$\& \ \text{a}_2=1, \ \text{b}_2=\frac{1}{2}, \ \text{c}_2=1 $
Since two lines are perpendicular, therefore
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$\Rightarrow1\times1=(-1)\times\frac{1}{2}+\frac{1}{\text{k}}\times(-1)=0$
$\Rightarrow1-\frac{1}{2}-\frac{1}{\text{k}}=0$
$\Rightarrow1-\frac{\text{k}-2}{2\text{k}}=0$
$\Rightarrow2\text{k}-\text{k}-2=0$
$\text{k}=2$
View full question & answer→Question 852 Marks
Write the equation of the plane parallel to the YOZ- plane and passing through (-4, 1, 0).
AnswerThe equation of the plane parallel ot the plane YOZ is x = b .....(i), where b is a constant. It is given that plane passes throught (-4, 1, 0). So, -4 = b
Substituting this value in (i), we get x = -4, which is the required equation of the plane.
View full question & answer→Question 862 Marks
Write the condition for the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ to be intersecting.
AnswerThe shortest distance d between the parallel lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
For the lines to be intersecting, d = 0.
$\Rightarrow\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}=0$
$\Rightarrow\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)$
View full question & answer→Question 872 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 882 Marks
Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y − z = 7.
AnswerLet the equation of a plane parallel to the given plane be
3x + 2y - z = k ....(1)
This passes through (2, -1, 1).
So, 3(2) + 2(-1) - (1) = k
k = 3
Substituting this in(1),
We get,
3x + 2y - z = 3, which is the equation of the required plane.
View full question & answer→Question 892 Marks
If a line makes angles of 90°, 60° and 30° with the positive direction of x, y, and z-axis respectively, find its direction cosines.
AnswerLet l, m and n be the direction cosines of a line.$\text{l}=\cos90^{\circ}=0$
$\text{m}=\cos60^{\circ}=\frac{1}{2}$ $\text{n}=\cos30^{\circ}=\frac{\sqrt{3}}{2}$ $\therefore$ The direction consines of the line are $0,\frac{1}{2},\frac{\sqrt{3}}{2}.$
View full question & answer→Question 902 Marks
What are the direction cosines of Y-axis?
AnswerThe y-axis makes angles 90°, 0° 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. 0, 1, 0.
View full question & answer→Question 912 Marks
If a line makes angles 90° and 60° respectively with the positive direction of x and y axes, find the angle which it makes with the positive direction of z-axis.
AnswerLet the direction cosines of the line be l, m and n.
We know that $l^2 + m^2 + n^2 = 1$
Let the line make angle $\theta$ with positive direction of the z-axis
$\alpha=90^\circ,\beta=60^\circ,\gamma=\theta$.
So, $\cos^290^\circ+\cos^260^\circ+\cos^2\theta=0$
$\Rightarrow0+\Big(\frac{1}{2}\Big)^2+\cos^2\theta=1$
$\Rightarrow\cos^2\theta=1-\frac{1}{4}$
$\Rightarrow\cos^2\theta=\frac{3}{4}$
$\Rightarrow\cos\theta=\pm\frac{\sqrt{3}}{4}$
$\Rightarrow\theta=30^\circ$or $150^\circ$
View full question & answer→Question 922 Marks
Write the direction consines of the line whose cartesian equations are 2x = 3y = -z.
AnswerWe have2x = 3y = -z
The equation of the given line can be re-written as
$\frac{\text{x}}{\frac{1}{2}}=\frac{\text{y}}{\frac{1}{3}}=\frac{\text{z}}{-1}$
$\frac{\text{x}}{3}=\frac{\text{y}}{2}=\frac{\text{z}}{-6}$
The diraction ratios of the line parallel to AB are proportional to 3, 2, -6.
Hence, the direction cosines of the line parallel to AB are proportional to
$\frac{3}{\sqrt{3^2+2^2+(-6)^2}},\frac{2}{\sqrt{3^2+2^2+(-6)^2}},\frac{-6}{\sqrt{3^2+2^2+(-6)^2}}$
$=\frac{3}{7},\frac{2}{7},-\frac{6}{7}$
View full question & answer→Question 932 Marks
Determine the direction cosines of the normal to the plane and the distance from the origin.
z = 2
AnswerThe equation of the plane is z = 2 or 0x + 0y + z = 2 ...(1)
The direction ratios of normal are 0, 0, and 1.
$\therefore\ \sqrt{0^2+0^2+1^2}=1$
Dividing both sides of equation (1) by 1, we obtain
0.x + 0.y + 1.z = 2
This 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 the perpendicular drawn from the origin.
Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.
View full question & answer→Question 942 Marks
Find the angle between the lines $\vec{\text{r}}=\big(2\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$ and $\vec{\text{r}}=7\hat{\text{i}}-6\hat{\text{k}}+\mu\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big).$
AnswerLet $\theta$ be the angle between the given lines. The given lines are parallel to the vectors $\vec{\text{b}}_1=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}},$ respectively.
So, the angle $\theta$ between the given lines is given by
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)}{\sqrt{3^2+2^2+6^2}\sqrt{1^2+2^+2^2}}$
$=\frac{3\times1+2\times2+6\times2}{\sqrt{49}\sqrt{9}}$
$=\frac{19}{21}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{19}{21}\big)$
Thus, the angle between the given lines is $\cos^{-1}\big(\frac{19}{21}\big).$
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