Question 13 Marks
Write the direction cosines of the line $\frac{\text{x}-2}{2}=\frac{2\text{y}-5}{-3},\text{z}=2.$
Answer We 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 23 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 33 Marks
A line passes throuth the point with position vector $2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and is in the direction of $3\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}.$ Find equations of the line in vector and cartesian form.
AnswerWe know that the vector equation of a line passing through a point with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}.$
Here,
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
So, the vector equation of the required line is
$\vec{\text{r}}=\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\big)$
Here, $\lambda$ is a parameter.
View full question & answer→Question 43 Marks
Show that the line through the points (1, -1, 2) and (3, 4, -2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).
AnswerSuppose vector $\vec{\text{a}}$ is passing through the points (1, -1, 2) and (3, 4, -2) and $\vec{\text{b}}$ passing through the points (0, 3, 2) and (3, 5, 6).
Then,
$\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
Now,
$\vec{\text{a}}.\vec{\text{b}}=\big(2\hat{\text{i}}+5\hat{\text{j}}-4\hat{\text{k}}\big).\big(3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)=0$
Hence, the given lines are perpendicular to each other.
View full question & answer→Question 53 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 $a_{1 }a_{2 }+ b_{1 }b_{2 }+ c_{1 }c_{2 }= 0$
$\therefore -3 (3k) + 2k \times 1 + 2 (-5) = 0$
$\Rightarrow -9k + 2k - 10 = 0$
$\Rightarrow 7k = - 10$
$\Rightarrow\text{k}=\frac{-10}{7}$
Therefore, for $\text{k}=-\frac{10}{7},$ the given lines are perpendicular to each other.
View full question & answer→Question 63 Marks
Find the vector equation of the line passing through the point A(1, 2, -1) and parallel to the line 5x - 25 = 14 - 7y = 35z.
AnswerThe equation of the line 5x - 25 = 14 - 7y = 35z can be re-written as
$\frac{\text{x}-5}{\frac{1}{5}}=\frac{\text{y}-2}{\frac{-1}{7}}=\frac{\text{z}}{\frac{1}{35}}$
$\Rightarrow\frac{\text{x}-5}{7}=\frac{\text{y}-2}{-5}=\frac{\text{z}}{1}$
Since the required line is parallel to the given line, so the direction ration of the required line are proportional to 7, -5, 1.
The vector equation of the required line passing through the point (1, 2, -1) and having direction ratios proportional to 7, -5, 1 is
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big).$
View full question & answer→Question 73 Marks
Show that the line joining the origin to the points (2, 1, 1) is perpendicular to the line detarmined by the points (3, 5, -1) and (4, 3, -1).
AnswerThe direction ratios of the line joining the origin to the point (2, 1, 1) are 2, 1, 1.
Let $\vec{\text{b}}_1=2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
The direction ratios ot the line joining the points (3, 5, -1) and (4, 3, -1) are 1, -2, 0.
Let $\vec{\text{b}}_2=\hat{\text{i}}-2\hat{\text{j}}+0\hat{\text{k}}$
Now,
$\vec{\text{b}}_1.\vec{\text{b}}_2=\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+0\hat{\text{k}}\big)$
$=2-2+0$.
$=0$
$\therefore\vec{\text{b}}_1\perp\vec{\text{b}}_2$
Hence, the two lines joining the given points are perpendicular to each other.
View full question & answer→Question 83 Marks
Write the value of $\lambda$ for which the lines $\frac{\text{x}-3}{-3}=\frac{\text{y}+2}{2\lambda}=\frac{\text{z}+4}{2}$ and $\frac{\text{x}+1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}+6}{-5}$ are perpendicular to each other.
Answer We have
$\frac{\text{x}-3}{-3}=\frac{\text{y}+2}{2\lambda}=\frac{\text{z}+4}{2}$
$\frac{\text{x}+1}{3\lambda}=\frac{\text{y}-2}{1}=\frac{\text{z}+6}{-5}$
The given lines are parallel to vector $\vec{\text{b}}_1=-3\hat{\text{i}}+2\lambda\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}_2=3\lambda\hat{\text{i}}+\hat{\text{j}}-5\hat{\text{k}}.$
For $\vec{\text{b}}_1\perp\vec{\text{b}}_2,$ we must have
$\vec{\text{b}}_1.\vec{\text{b}}_2=0$
$\Rightarrow\big(-3\hat{\text{i}}+2\lambda\hat{\text{j}}+2\hat{\text{k}}\big).\big(3\lambda\hat{\text{i}}+\hat{\text{j}}-5\hat{\text{k}}\big)=0$
$\Rightarrow-7\lambda-10=0$
$\Rightarrow\lambda=-\frac{10}7{}$
View full question & answer→Question 93 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 other.
AnswerThe direction ratios of 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 proportional to 7, -5, 1 and 1, 2, 3, respectiveiy.
Let:
$\vec{\text{b}}_1=7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Now,
$\vec{\text{b}}_1.\vec{\text{b}}_2=\big(7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=7-10+3$
$=0$
$\therefore\vec{\text{b}}_1\perp\vec{\text{b}}_2$
Hence, the given lines are perpendicular to each other.
View full question & answer→Question 103 Marks
Write the direction cosines of the line whose cartesian equations are 6x -2 = 3y + 1 =2z - 4.
AnswerWe have
6x -2 = 3y + 1 =2z - 4
The equation of given line can be re-written as
$\frac{\text{x}-\frac{1}{3}}{\frac{1}{6}}=\frac{\text{y}+\frac{1}{3}}{\frac{1}{3}}=\frac{\text{z}-2}{\frac{1}{2}}$
$\Rightarrow\frac{\text{x}-\frac{1}{3}}{1}=\frac{\text{y}+\frac{1}{3}}{2}=\frac{\text{z}-2}{3}$
The direction ratios of the line parallel to AB are proportional to 1, 2, 3.
Hence, the direction cosines of the line parallel to AB are proportional to
$\frac{1}{\sqrt{1^2+2^2+3^2}},\frac{2}{\sqrt{1^2+2^2+3^2}},\frac{3}{\sqrt{1^2+2^2+3^2}}$
$=\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$
View full question & answer→Question 113 Marks
Write the angle between the lines $\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}-2}{1}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}}{2}=\frac{\text{z}-1}{3}.$
AnswerWe have
$\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}-2}{1}$
$\frac{\text{x}-1}{1}=\frac{\text{y}}{2}=\frac{\text{z}-1}{3}$
The given lines are parallel to the vectors $\vec{\text{b}}_1=7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Let $\theta$ be the angle between the given lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)}{\sqrt{7^2+(-5)^2+1^2}\sqrt{1^2+2^2+3^2}}$
$=\frac{7-10+3}{\sqrt{49+25+1}\sqrt{1+4+9}}$
$=0$
View full question & answer→Question 123 Marks
Find the value of $\lambda$ so that the following lines are perpendicular to each other.$\frac{\text{x}-5}{5\lambda+2}=\frac{2-\text{y}}{5}=\frac{1-\text{z}}{-1},\frac{\text{x}}{1}=\frac{2\text{y}+1}{4\lambda}=\frac{1-\text{z}}{-3}$
AnswerThe equation of the given lines $\frac{\text{x}-5}{5\lambda+2}=\frac{2-\text{y}}{5}=\frac{1-\text{z}}{-1}$ and $\frac{\text{x}}{1}=\frac{2\text{y}+1}{4\lambda}=\frac{1-\text{z}}{-3}$ can be re-written as $\frac{\text{x}-5}{5\lambda+2}=\frac{\text{y}-2}{-5}=\frac{\text{z}-1}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}+\frac{1}{2}}{2\lambda}=\frac{\text{z}-1}{3}$
Since the given lines are pependicular to each other, we have
$(5\lambda+2)1-5(2\lambda)+1(3)=0$
$\Rightarrow5\lambda=5$
$\Rightarrow\lambda=1$
View full question & answer→Question 133 Marks
Find in vector form as wel as in cartesian form, the equation of the line passing through the points $A(1, 2, -1)$ and $B(2, 1, 1).$
AnswerWe know that, equation of line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is
$\frac{\text{x}-\text{x}_1}{\text{x}_2-\text{x}_1}=\frac{\text{y}-\text{y}_1}{\text{y}_2-\text{y}_1}=\frac{\text{z}-\text{z}_1}{\text{z}_2-\text{z}_1}\dots(1)$
Here, $\big(\text{x}_1,\text{y}_1,\text{z}_1\big)=\text{A}(1,2,-1)$
$\big(\text{x}_2,\text{y}_2,\text{z}_2\big)=\text{B}(2,1,1)$
Using equation $(1),$ equation of line $AB$
$\frac{\text{x}-1}{2-1}=\frac{\text{y}-2}{1-2}=\frac{\text{z}+1}{1+1}$
$\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}+1}{2}=\lambda ($Say$)$
$\text{x}=\lambda+1,\text{y}=-\lambda+2,\text{z}=2\lambda-1$
vector form of equation of line $Ab$ is,
$\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=(\lambda+1)\hat{\text{i}}+(-\lambda+2)\hat{\text{j}}+(2\lambda-1)\hat{\text{k}}$
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$
View full question & answer→Question 143 Marks
Find the points on the line $\frac{\text{x}+2}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-3}{2}$ at a distance of 5 units from the point P(1, 3, 3).
AnswerThe coordinates of any point on the line $\frac{\text{x}+2}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-3}{2}$ are given by
$\frac{\text{x}+2}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-3}{2}=\lambda$
$\Rightarrow\text{x}=3\lambda-2,\text{y}=2\lambda-1,\text{z}=2\lambda+3\dots(1)$
Let the coordinates of the desired point be $(3\lambda-2,2\lambda-1,2\lambda+3)$
The distance between this point and (1, 3, 3) is 5 units.
$\therefore\sqrt{(3\lambda-2-1)^2+(2\lambda-1-3)^2+(2\lambda+3-3)^2}=5$
$\Rightarrow(3\lambda-3)^2+(2\lambda-4)^2+(2\lambda)^2=25$
$\Rightarrow17\lambda^2-34\lambda=0$
$\Rightarrow\lambda(\lambda-2)=0$
$\Rightarrow\lambda=0$ or $2$
Substituting the values of $\lambda$ in (1), we get the coordinates of the desired point as (-2, -1, 3) and (4, 3, 7).
View full question & answer→Question 153 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 163 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 173 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 183 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 193 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 203 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 213 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 223 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 233 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
$\frac{\sqrt{3}}{\sqrt{(\sqrt{3})^2+4^2+6^2}},\frac{4}{\sqrt{(\sqrt{3})^2+4^2+6^2}},\frac{6}{\sqrt{(\sqrt{3})^2+4^2+6^2}}$
$=\frac{\sqrt{3}}{\sqrt{55}},\frac{4}{\sqrt{55}},\frac{6}{\sqrt{55}}$
View full question & answer→Question 243 Marks
Find the cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by $\frac{\text{x}+3}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}+8}{6}.$
AnswerWe know that the cartesian equation of a line passing with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ is $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_2}{\text{b}}=\frac{\text{z}-\text{z}_3}{\text{c}}.$
Here,
$\vec{\text{a}}=-2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}$
The cartesian equation of the required line is
$\frac{\text{x}-(-2)}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}-(-5)}{6}$
$=\frac{\text{x}+2}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}+5}{6}$
View full question & answer→Question 253 Marks
Find the vector equation of the line passing through the point (2, -1, -1) which is parallel to the line 6x - 2 = 3y +1 =2z - 2.
AnswerThe equation of the line 6x - 2 = 3y + 1 = 2z - 2 can be re-written as
$\frac{\text{x}-\frac{1}{3}}{\frac{1}{6}}=\frac{\text{y}+\frac{1}{3}}{\frac{1}{3}}=\frac{\text{z}-1}{\frac{1}{2}}$
$=\frac{\text{x}-\frac{1}{3}}{1}=\frac{\text{y}+\frac{1}{3}}{2}=\frac{\text{z}-1}{3}$
Since the reqired line is parallel to the given line, the direction ratios of the recuired line are proportional to 1, 2, 3.
The vector equation of the required line passing through the point (2, -1, -1) and having direction ratios proportional to 1, 2, 3 is $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big).$
View full question & answer→Question 263 Marks
Find the perpendicular distence of the point (3, -1, 11) from the line $\frac{\text{x}}{2}=\frac{\text{y}-2}{-3}=\frac{\text{z}-3}{4}.$
AnswerLet the point (3, -1, 11) be P and the point through which the line passes be Q (0, 2, 3).
The line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
Now,
$\therefore\vec{\text{b}}\times\overrightarrow{\text{PQ}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-3&4\\-3&3&-8\end{vmatrix}$
$=12\hat{\text{i}}+4\hat{\text{j}}+15\hat{\text{k}}$
$\Rightarrow|\vec{\text{b}}\times\overrightarrow{\text{PQ}}|=\sqrt{12^2+4^2+15^2}$
$=\sqrt{144+16+225}$
$=\sqrt{385}$
$\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}$
$=\frac{\sqrt{385}}{\sqrt{29}}$
$=\frac{\sqrt{385}}{\sqrt{29}}$
View full question & answer→Question 273 Marks
Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the other one is obtained by joining the points (3, 1, 4) and (7, 2, 12).
AnswerThe direction ratios of the line joining the points (3, 1, 4) and (7, 2, 12) are proportional to 4, 1, 8.
Let $\vec{\text{m}}_1$ and $\vec{\text{m}}_2$ be vectors parallel to the lines having direction ratios proportional to 2, 2, 1 and 4, 1, 8.
Now,
$\vec{\text{b}}_1=2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}_2=4\hat{\text{i}}+\hat{\text{j}}+8\hat{\text{k}}$
If $\theta$ is the angle between the given lines, then
$\cos\theta=\frac{\vec{\text{m}}_1.\vec{\text{m}}_2}{|\vec{\text{m}}_1||\vec{\text{m}}_2|}$
$=\frac{\big(2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big).\big(4\hat{\text{i}}+\hat{\text{j}}+8\hat{\text{k}}\big)}{\sqrt{2^2+2^2+1^2}\sqrt{4^2+1^2+8^2}}$
$=\frac{8+2+8}{3\times9}$
$=\frac{2}{3}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{2}{3}\big)$
View full question & answer→Question 283 Marks
Find the vector equation of a line passing through (2, -1, 1) and parallel to the line whose equations are $\frac{\text{x}-3}{2}=\frac{\text{y}+1}{7}=\frac{\text{z}-2}{-3}.$
AnswerWe know that the vector equation of a line passing through a point with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
Here,
$\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}$
Vector equation of the required line is
$\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}\big)$
Here, $\lambda$ is a parameter.
View full question & answer→Question 293 Marks
Write the angle between the lines 2x = 3y = -z and 6x = -y = -4z.
AnswerWe have
2x = 3y = -z
6x = -y = -4z
The given lines can be re-written as
$\frac{\text{x}}{\frac{1}{2}}=\frac{\text{y}}{\frac{1}{3}}=\frac{\text{z}}{-1}$ and $\frac{\text{x}}{\frac{1}{6}}=\frac{\text{y}}{-1}=\frac{\text{z}}{-\frac{1}{4}}$
$\Rightarrow\frac{\text{x}}{3}=\frac{\text{y}}{2}=\frac{\text{z}}{-6}$ and $\frac{\text{x}}{2}=\frac{\text{y}}{-12}=\frac{\text{z}}{-3}$
These lines are parallel to vectors $\vec{\text{b}}_1=3\hat{\text{i}}+2\hat{\text{j}}-6\hat{\text{k}}$ and $\vec{\text{b}}_2=2\hat{\text{i}}-12\hat{\text{j}}-3\hat{\text{k}}$
Let $\theta$ be the angle between these lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(3\hat{\text{i}}+2\hat{\text{j}}-6\hat{\text{k}}\big).\big(2\hat{\text{i}}-12\hat{\text{j}}-3\hat{\text{k}}\big)}{\sqrt{3^2+2^2+(-6)^2}\sqrt{2^2+(-12)^2+(-3)^2}}$
$=\frac{6-24+18}{\sqrt{9+4+36}\sqrt{4+144+9}}$
$=0$
$\Rightarrow\theta=\frac{\pi}{2}$
View full question & answer→Question 303 Marks
Find the length of the perpendicular drow from the point (5, 4, -1) to the line $\vec{\text{r}}=\hat{\text{i}}+\lambda\big(2\hat{\text{i}}+9\hat{\text{j}}+5\hat{\text{k}}\big).$
AnswerLet the point (5, 4, -1) be P and the point through which the line passes be Q(1, 0, 0).The line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+9\hat{\text{k}}+5\hat{\text{k}}.$
Now,
$\therefore\vec{\text{b}}\times\overrightarrow{\text{PQ}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&9&5\\-4&-4&1\end{vmatrix}$
$=29\hat{\text{i}}-22\hat{\text{j}}+28\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|=\sqrt{29^2+(-22)^2+28^2}$
$=\sqrt{841+484+784}$
$=\sqrt{2109}$
$\big|\vec{\text{b}}\big|=\sqrt{2^+9^2+5^2}$
$=\sqrt{4+81+25}$
$=\sqrt{110}$
$\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}$
$=\frac{\sqrt{2109}}{\sqrt{110}}$
View full question & answer→