Question 11 Mark
The Cartesian equation of the line passing through the point $(1,-3,2)$ and parallel to the line $\vec{r}=(2+\lambda) \hat{i}+\lambda \hat{j}+(2 \lambda-1) \hat{k}$ is
AnswerGiven line is
$\begin{aligned}
\vec{r} & =(2+\lambda) \hat{i}+\lambda \hat{j}+(2 \lambda-1) \hat{k} \\
& =2 \hat{i}-\hat{k}+\lambda(\hat{i}+\hat{j}+2 \hat{k})
\end{aligned}$
And, position vector of a point on the required line is
$\hat{i}-3 \hat{j}+2 \hat{k} \text {. }
$Thus, vector equation of the line is
$\vec{r}=(\hat{i}-3 \hat{j}+2 \hat{k})+\lambda(\hat{i}+\hat{j}+2 \hat{k}), \lambda \in R
$The line passes through $(1,-3,2)$ and has direction ratios $1,1,2$
$\therefore \quad$ The cartesian equation of the line is
$\frac{x-1}{1}=\frac{y+3}{1}=\frac{z-2}{2}$
View full question & answer→Question 21 Mark
The lines $\frac{1-x}{2}=\frac{y-1}{3}=\frac{z}{1}$ and $\frac{2 x-3}{2 p}=\frac{y}{-1}=\frac{z-4}{7}$ are perpendicular to each other for $p$ equal to :
AnswerAs the given lines can be written as
$\frac{x-1}{-2}=\frac{y-1}{3}=\frac{z-0}{1} \text { and } \frac{x-3 / 2}{p}=\frac{y}{-1}=\frac{z-4}{7}$
These lines are perpendicular, then
$
\begin{aligned}
& -2 \times p+3(-1)+1 \times 7=0 \Rightarrow-2 p-3+7=0 \\
\Rightarrow \quad & -2 p+4=0 \Rightarrow-2 p=-4 \therefore p=2
\end{aligned}$
View full question & answer→Question 31 Mark
The coordinates of the foot of the perpendicular drawn from the point $(0,1,2)$ on the $x$-axis are given by:
View full question & answer→Question 41 Mark
If a line makes an angle of $30^{\circ}$ with the positive direction of $x$-axis, $120^{\circ}$ with the positive direction of $y-$axis, then the angle which it makes with the positive direction of $z-$axis is :
AnswerLet the angle made with positive direction of $z$-axis be $\gamma$.
Then, $\cos ^2 30^{\circ}+\cos ^2 120^{\circ}+\cos ^2 \gamma=1$
$\Rightarrow\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{-1}{2}\right)^2+\cos ^2 \gamma=1$
$\Rightarrow \cos ^2 \gamma=1-\frac{3}{4}-\frac{1}{4}=0 \Rightarrow \gamma=90^{\circ}$
View full question & answer→Question 51 Mark
The vector equation of a line passing through the point $(1,-1,0)$ and parallel to $Y$-axis is :
AnswerEquation of line passing through the point $(1,-1,0)$ and parallel to $y$ - axis is given by
$\vec{r}=(\hat{i}-\hat{j}+0 \hat{k})+\lambda(0 \hat{i}+\hat{j}+0 \hat{k}) \quad \therefore \quad \vec{r}=(\hat{i}-\hat{j})+\lambda \hat{j}$
View full question & answer→Question 61 Mark
The angle which the line $\frac{x}{1}=\frac{y}{-1}=\frac{z}{0}$ makes with the positive direction of $Y$-axis is :
AnswerGiven, $\frac{x}{1}=\frac{y}{-1}=\frac{z}{0}$.........(i)
Dr's of (i) is $1,-1,0$.
Dr's of $y$-axis is $0,1,0$.
Now, $\cos \theta=\left|\frac{1 \times 0+1(-1)+0 \times 0}{\sqrt{1^2+(-1)^2} \sqrt{1^2}}\right|=\frac{1}{\sqrt{2}} \quad \therefore \quad \theta=\frac{7 \pi}{4}$
View full question & answer→Question 71 Mark
Direction ratios of a vector parallel to line $\frac{x-1}{2}=-y=\frac{2 z+1}{6}$ are :
AnswerThe given line can be written as
$
\frac{x-1}{2}=\frac{y}{-1}=\frac{z+1 / 2}{3}
$
So, direction ratios of line parallel to given line is <2,-1,3>
View full question & answer→Question 81 Mark
The angle between the lines $2 x=3 y=-z$ and $6 x=-y=-4 z$ is
AnswerThe given equation of lines can be rewritten as
$\frac{x-0}{1 / 2}=\frac{y-0}{1 / 3}=\frac{z-0}{-1} \text { and } \frac{x-0}{1 / 6}=\frac{y-0}{-1}=\frac{z-0}{-1 / 4}$
$\therefore a_1=\frac{1}{2}, b_1=\frac{1}{3}, c_1=-1 \text { and } a_2=\frac{1}{6}, b_2=-1, c_2=\frac{-1}{4}$
Now, $\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$
$=\frac{\frac{1}{2} \cdot \frac{1}{6}+\frac{1}{3} \cdot(-1)+(-1) \cdot\left(\frac{-1}{4}\right)}{\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^2+(-1)^2} \sqrt{\left(\frac{1}{6}\right)^2+(-1)^2+\left(\frac{-1}{4}\right)^2}}=0$
$\Rightarrow \cos \theta=0 \Rightarrow \theta=90^{\circ}$
View full question & answer→Question 91 Mark
A line $\overrightarrow{O P}$ in space, represented by the figure below; has a magnitude of $2 \sqrt{2}$ units.
Which of these are the direction ratios of $\overrightarrow{O P}$ ? Answer$(\sqrt{2}, 2, \sqrt{2})$
View full question & answer→Question 101 Mark
Direction cosines of the line $\frac{x-1}{2}=\frac{1-y}{3}=\frac{2 z-1}{12}$ are:
AnswerGiven equation of the line is $\frac{x-1}{2}=\frac{1-y}{3}=\frac{2 z-1}{12}$ Which can be written as
$\frac{x-1}{2}=\frac{(y-1)}{-3}=\frac{z-\frac{1}{2}}{6}$
$\therefore \quad$ The direction cosines are
$\\frac{2}{\sqrt{(2)^2+(-3)^2+(6)^2}}, \frac{-3}{\sqrt{(2)^2+(-3)^2+(6)^2}}, \frac{6}{\sqrt{(2)^2+(-3)^2+(6)^2}}$
$=\frac{2}{\sqrt{49}}, \frac{-3}{\sqrt{49}}, \frac{6}{\sqrt{49}}=\frac{2}{7}, \frac{-3}{7}, \frac{6}{7}$
View full question & answer→Question 111 Mark
The vector equation of a line which passes through the point $(2,-4,5)$ and is parallel to the line $\frac{x+3}{3}=\frac{4-y}{2}=\frac{z+8}{6}$ is :
AnswerLet $\vec{a}$ be the position vector of the point $(2,-4,5)$, then $\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k}$.
The given equation of line is $\frac{x+3}{3}=\frac{4-y}{2}=\frac{z+8}{6}$ $\Rightarrow \frac{x+3}{3}=\frac{y-4}{-2}=\frac{z+8}{6}$
$\therefore \quad$ Direction ratios of (i) are $3,-2,6$.
Let $\vec{b}$ be the vector parallel to line (i).
Then, $\vec{b}=3 \hat{i}-2 \hat{j}+6 \hat{k}$
$\therefore$ The vector equation of required line is $\vec{r}=\vec{a}+\lambda \vec{b}$ $\Rightarrow \vec{r}=(2 \hat{i}-4 \hat{j}+5 \hat{k})+\lambda(3 \hat{i}-2 \hat{j}+6 \hat{k})$
View full question & answer→Question 121 Mark
If the direction cosines of a line are $\left(\frac{1}{a}, \frac{1}{a}, \frac{1}{a}\right)$, then
AnswerGiven that the direction cosines of a line are
$\left(\frac{1}{a}, \frac{1}{a}, \frac{1}{a}\right)$
We know that the sum of squares of the direction cosines is 1 .
$\Rightarrow \frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{a^2}=1 \Rightarrow \frac{3}{a^2}=1 \Rightarrow a^2=3$
$\Rightarrow a= \pm \sqrt{3}$
View full question & answer→Question 131 Mark
If a line makes angles of $90^{\circ}, 135^{\circ}$ and $45^{\circ}$ with the $x, y$ and $z$ axes respectively, then its direction cosines are
AnswerDirection cosines are $\left\langle\cos 90^{\circ}, \cos 135^{\circ}, \cos 45^{\circ}\right\rangle$
$
=\left\langle 0, \cos \left(90^{\circ}+45^{\circ}\right), \frac{1}{\sqrt{2}}\right\rangle=\left\langle 0,-\sin 45^{\circ}, \frac{1}{\sqrt{2}}\right\rangle=\left\langle 0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right\rangle$
View full question & answer→Question 141 Mark
Distance of the point $(p, q, r)$ from $y$-axis is
AnswerGiven point is $(p, q, r)$
The foot of perpendicular drawn from point $(p, q, r)$ on the $y$-axis is $(0, q, 0)$.
Now, distance between these two points is
$\sqrt{(p-0)^2+(q-q)^2+(r-0)^2}=\sqrt{p^2+r^2}$
View full question & answer→Question 151 Mark
$P$ is a point on the line joining the points $A(0,5,-2)$ and $B(3,-1,2)$. If the $x$-coordinate of $P$ is 6 , then its $z$-coordinate is
AnswerThe line through the points $(0,5,-2)$ and $(3,-1,2)$ is
$\frac{x}{3-0}=\frac{y-5}{-1-5}=\frac{z+2}{2+2} \text { or } \frac{x}{3}=\frac{y-5}{-6}=\frac{z+2}{4}$
Any point on the line is $P(3 k,-6 k+5,4 k-2)$, where $k$ is an arbitrary scalar.
$\because \quad 3 k=6 \Rightarrow k=2$
The $z$-coordinate of the point $P$ will be $4 \times 2-2=6$.
View full question & answer→Question 161 Mark
A line $m$ passes through the point $(-4,2,-3)$ and is parallel to line $n$, given by:
$\frac{-x-2}{4}=\frac{y+3}{-2}=\frac{2 z-6}{3}$
The vector equation of line $m$ is given by: $\vec{r}=(-4 \hat{i}+2 \hat{j}-3 \hat{k})+\lambda(p \hat{i}+q \hat{j}+r \hat{k})$, where $\lambda \in R$
Which of the following could be the possible values for $p, g$ and $r$ ?
View full question & answer→Question 171 Mark
If the two lines $L_1: x=5, \frac{y}{3-\alpha}=\frac{z}{-2} \quad L_2: x=2, \frac{y}{-1}=\frac{z}{2-\alpha}$ are perpendicular, then the value of $\alpha$ is
AnswerThe given lines are perpendicular, if
$a_1 a_2+b_1 b_2+c_1 c_2=0$
Here, $L_1: \frac{x-5}{0}=\frac{y-0}{3-\alpha}=\frac{z-0}{-2}$
$L _2: \frac{x-2}{0}=\frac{y-0}{-1}=\frac{z-0}{2-\alpha}$
Here, $a_1, b_1, c_1$ are $0,3-\alpha,-2$, and $a_2, b_2, c_2$ are $0,-1,2-\alpha$ respectively.
$\begin{array}{ll}
\therefore & 0 \times 0-(3-\alpha)-2(2-\alpha)=0 \\
\Rightarrow & \alpha=\frac{7}{3}
\end{array}$
View full question & answer→Question 181 Mark
The length of the perpendicular drawn from the point $(4,-7,3)$ on the $y$-axis is
AnswerLet $P(4,-7,3)$ be the given point and $A$ be a point on $y$-axis s.t. $P A \perp$ to $y$-axis.
$\therefore A \equiv(0,-7,0)$
$\text { Now, }
P A =\sqrt{(4-0)^2+(-7-(-7))^2+(3-0)^2}$
$ =\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5 \text { units }$
View full question & answer→Question 191 Mark
The vector equation of $X Y$-plane is
AnswerVector equation of $X Y$-plane is $\vec{r} \cdot \hat{k}=0$.
View full question & answer→Question 201 Mark
Direction cosines of the line that makes equal angles with the three axes in space are
Answer(c) : Since $l=m=n$ and $l^2+m^2+n^2=1$
$\therefore \quad l=m=n= \pm \frac{1}{\sqrt{3}}$.
View full question & answer→Question 211 Mark
If lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-5}{1}$ $=\frac{z-6}{-5}$ are mutually perpendicular, then $k$ is equal to
Answer(a) : Lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are perpendicular if $a_1 a_2+b_1 b_2+c_1 c_2=0$.
$
\Rightarrow-3(3 k)+2 k+2(-5)=0 \Rightarrow k=-\frac{10}{7}
$
View full question & answer→Question 221 Mark
The equation of the line joining the points $(-3,4,11)$ and $(1,-2,7)$ is
Answer(b) : DR's of the line joining the given points are
$
\{1-(-3),-2-4,7-11\}
$
i.e., $(4,-6,-4)$ or $(-2,3,2)$
Now, Equation of line passing through $(-3,4,11)$ and having direction ratios $-2,3,2$ is $\frac{x+3}{-2}=\frac{y-4}{3}=\frac{z-11}{2}$
View full question & answer→Question 231 Mark
length of the $1^{er}$ from the point $(0, -1, 3)$ to the plane $2x + y - 2z + 1 = 0$ is:
View full question & answer→Question 241 Mark
The direction ratios of the normal to the plane 7x + 4y - 2z + 5 = 0 are:
Question 251 Mark
If the two lines $L_1: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$
$L_2: x=2, \frac{y}{-1}=\frac{z}{2-\alpha}$ are perpendicular, then the value of $\alpha$ is
Answer(d) : The given lines are perpendicular, if
$
a_1 a_2+b_1 b_2+c_1 c_2=0 ...(i)
$
Here, $L_1: \frac{x-5}{0}=\frac{y-0}{3-\alpha}=\frac{z-0}{-2}$
$
L_2: \frac{x-2}{0}=\frac{y-0}{-1}=\frac{z-0}{2-\alpha}
$
Here, $a_1, b_1, c_1$ are $0,3-\alpha,-2$, and $a_2, b_2, c_2$ are $0,-1$, $2-\alpha$ respectively.
$
\begin{array}{ll}
\therefore & 0 \times 0-(3-\alpha)-2(2-\alpha)=0 \\
\Rightarrow & \alpha=\frac{7}{3} & [From (i)]
\end{array}
$
View full question & answer→Question 261 Mark
The equation of a line passing through the point $(-3,2,-4)$ and equally inclined to the axes are
Answer$(b) :$ Since, line equally inclined to the axes.
$\therefore l=m=n ...(i)$
The required equation of line is
$\frac{x+3}{l}=\frac{y-2}{l}=\frac{z+4}{l} \text { [Using (i)] }$
$\Rightarrow \frac{x+3}{1}=\frac{y-2}{1}=\frac{z+4}{1} \Rightarrow x+3=y-2=z+4$
View full question & answer→Question 271 Mark
The cartesian equation of a line is $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$. The vector equation for the line is
Answer(b) : The given cartesian equation is
$
\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2} \text {. }
$
The line passes through the point $(-3,5,-6)$ and is parallel to vector $2 \hat{i}+4 \hat{j}+2 \hat{k}$.
Hence, the vector equation of the line is
$
\vec{r}=-3 \hat{i}+5 \hat{j}-6 \hat{k}+\lambda(2 \hat{i}+4 \hat{j}+2 \hat{k}) .
$
View full question & answer→Question 281 Mark
What are the direction ratios of normal to the plane 2x - y + 2z + 1 = 0:
Answer
- (2, -1, 2)
Solution:
Direction ratios of normal to the plane ax + by + cz + d = 0, are
a, b, c. So, here in the question the given plane is 2x - y + 2z + 1
= Thus, the direction ratios are 2, -1, 2
View full question & answer→Question 291 Mark
Can $\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}$ be the direction cosines of any directed line:
Answer
- No
Solution:
No, they can not be the direction cosines of any directed line. As the sum of square of them is not 1.As
$=\Big(\frac{1}{\sqrt{3}}\Big)^2+\Big(\frac{2}{\sqrt{3}}\Big)^2+\Big(\frac{-2}{\sqrt{3}}\Big)^2=\frac{1+4+4}{3}=3$ View full question & answer→Question 301 Mark
The angle between the planes 2x - y + z = 6 and x + y + 2z = 7 is:
Question 311 Mark
$l = m = n = 1$ represents the direction cosines of:
AnswerSuppose, $l, m, n$ are direction cosines
$\Rightarrow 1^2 + m^2 + n^2 = 1$
But $1 = m = n = 1$
$\Rightarrow 3m^2 = 1$
$\Rightarrow 1 = m = n = \frac{1}{\sqrt3}$
which are not direction cosines of either of the three co-ordinate axes.
View full question & answer→Question 321 Mark
If the diraction ratios of a line are proportional to 1, -3, 2, then its diraction cosines are:
Answer
- $\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
Solution:
The diraction ratios of the line are proportional to 1, -3, 2.
$\therefore$ The direction cosines of the line are
$\frac{1}{\sqrt{1^2+(-3)^2+2^2}},\frac{-3}{\sqrt{1^2+(-3)^2+2^2}},\frac{2}{\sqrt{1^2+(-3)^2+2^2}}$
$=\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$ View full question & answer→Question 331 Mark
The direction coisines of the y-axis are:
Question 341 Mark
The eqution of the plane which cute equal intercepts of unit length on the coordinate axes is:
Answer
- x + y + z = 1
Solution:
We know that the equation of aplane whose intercepts are a, b, c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ....(1)$
It is given that a = b = c
So, from (1),
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
$\Rightarrow\text{x}+\text{y}+\text{z}=\text{a}\ ....(2)$
Since it is given that the intercepts of the required plane are of unit length,
a = b = c = 1
Substituting a = 1 in (2), we get
x + y + z = 1 View full question & answer→Question 351 Mark
What are the DRs of vector parallel to (2, -1, 1) and (3, 4, -1):
Answer
- (1, 5, -2)
Solution:
Required DRs are (3 - 2, 4 + 1, -1 - 1) ie, (1, 5, -2)
View full question & answer→Question 361 Mark
If a line makes angles $\alpha,\beta,\gamma$ with the axis then $\cos 2\alpha+ \cos 2\beta +\cos 2\gamma=$
View full question & answer→Question 371 Mark
The equation $x^2- x - 2 = 0$ in three dimensional space is represented by:
AnswerA pair of parallel planes
View full question & answer→Question 381 Mark
If the projections of the line segment AB on the coordinate axes are 2, 3, 6, then the square of the sine of the angle made by AB with x = 0, is:
Question 391 Mark
The length of the perpendicular drawn from the point $(4,-7,3)$ on the $y$-axis is
Answer(c) : Let $P(4,-7,3)$ be the given point and $A$ be a point on $y$-axis s.t. $P A \perp$ to $y$-axis.
$
\therefore A \equiv(0,-7,0)
$
Now, $P A=\sqrt{(4-0)^2+(-7-(-7))^2+(3-0)^2}$
$
=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5 \text { units }
$
View full question & answer→Question 401 Mark
$P$ is a point on the line joining the points $A(0$, $5,-2)$ and $B(3,-1,2)$. If the $x$-coordinate of $P$ is 6 , then its $z$-coordinate is
Answer(b) : The line through the points $(0,5,-2)$ and $(3,-1,2)$ is
$
\frac{x}{3-0}=\frac{y-5}{-1-5}=\frac{z+2}{2+2} \text { or } \frac{x}{3}=\frac{y-5}{-6}=\frac{z+2}{4}
$
Any point on the line is $P(3 k,-6 k+5,4 k-2)$, where $k$ is an arbitrary scalar.
$
\because 3 k=6 \Rightarrow k=2
$
The $z$-coordinate of the point $P$ will be $4 \times 2-2=6$.
View full question & answer→Question 411 Mark
If the direction cosines of a line are $\left(\frac{1}{a}, \frac{1}{a}, \frac{1}{a}\right)$, then
Answer(d) : Given that the direction cosines of a line are
$
\left(\frac{1}{a}, \frac{1}{a}, \frac{1}{a}\right) \text {. }
$
We know that the sum of squares of the direction cosines is 1 .
$
\Rightarrow \frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{a^2}=1 \Rightarrow \frac{3}{a^2}=1 \Rightarrow a^2=3 \Rightarrow a= \pm \sqrt{3}
$
View full question & answer→Question 421 Mark
If the line joining $(2,3,-1)$ and $(3,5,-3)$ is perpendicular to the line joining $(1,2,3)$ and $(3,5, \lambda)$, then $\lambda=$
Answer(d) : DR's of the given lines are 1, 2, -2 and 2, 3, $\lambda-3$.
Since, lines are perpendicular. $\therefore a_1 a_2+b_1 b_2+c_1 c_2=0$
$
\Rightarrow \quad 1 \times 2+2 \times 3-2(\lambda-3)=0 \Rightarrow \lambda=7
$
View full question & answer→Question 431 Mark
The value of $p$, so that the lines $\frac{1-x}{3}=\frac{7 y-14}{2 p}$ $=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ intersect at right angle, is
Answer(b) : Equation of the given lines can be written in the standard form as
$
\frac{x-1}{-3}=\frac{y-2}{\frac{2 p}{7}}=\frac{z-3}{2} \text { and } \frac{x-1}{-\frac{3 p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}
$
$\because \quad$ Lines are perpendicular to each other.
$\therefore \quad a_1 a_2+b_1 b_2+c_1 c_2=0$
$
\Rightarrow(-3)\left(\frac{-3 p}{7}\right)+\left(\frac{2 p}{7}\right)(1)+(2)(-5)=0 \Rightarrow p=\frac{70}{11}
$
View full question & answer→Question 441 Mark
If a line makes angles $\frac{\pi}{2}, \frac{3 \pi}{4}$ and $\frac{\pi}{4}$ with $X, Y$, and $Z$-axes respectively, then its direction cosines are
Answer(a) : Here, $l=\cos \frac{\pi}{2}=0$
$
m=\cos \frac{3 \pi}{4}=\cos \left(\pi-\frac{\pi}{4}\right)=-\cos \frac{\pi}{4}=\frac{-1}{\sqrt{2}}
$
and $n=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$
$\therefore \quad$ Direction cosines are $0, \frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.
View full question & answer→Question 451 Mark
Distance of the point $(\alpha, \beta, \gamma)$ from $y$-axis is
Answer(d) : Foot of perpendicular from $(\alpha, \beta, \gamma)$ on the $y$-axis is $(0, \beta, 0)$.
$\therefore \quad$ Distance of $(\alpha, \beta, \gamma)$ from $y$-axis $=$ distance of $(\alpha, \beta, \gamma)$ from $(0, \beta, 0)$
$
=\sqrt{(0-\alpha)^2+(\beta-\beta)^2+(0-\gamma)^2}=\sqrt{\alpha^2+\gamma^2}
$
View full question & answer→Question 461 Mark
Find the direction cosines of the line $\frac{x-2}{2}=\frac{2 y-5}{-3}=\frac{z+1}{0}$.
Answer$(d) :$ The given line is
$\frac{x-2}{2}=\frac{2 y-5}{-3}=\frac{z+1}{0}$
$\Rightarrow \frac{x-2}{2}=\frac{y-5 / 2}{-3 / 2}=\frac{z+1}{0}$
Direction cosines are
$\frac{2}{\sqrt{2^2+\left(\frac{-3}{2}\right)^2+0^2}}, \frac{-3 / 2}{\sqrt{2^2+\left(\frac{-3}{2}\right)^2+0^2}}, \frac{0}{\sqrt{2^2+\left(\frac{-3}{2}\right)^2+0^2}}$
$\text { i.e., } \frac{2}{5 / 2}, \frac{-3 / 2}{5 / 2}, 0 \text { i.e., } \frac{4}{5}, \frac{-3}{5}, 0$
View full question & answer→Question 471 Mark
A line makes angles $\alpha, \beta$ and $\gamma$ with the co-ordinate axes. If $\alpha+\beta=90^{\circ}$, then the value of angle $\gamma$ is
Answer$\text { (b): We know that } \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
$\Rightarrow \cos ^2 \alpha+\cos ^2\left(90^{\circ}-\alpha\right)+\cos ^2 \gamma=1 \quad\left[\because \alpha+\beta=90^{\circ}\right]$
$\Rightarrow \cos ^2 \alpha+\sin ^2 \alpha+\cos ^2 \gamma=1$
$\Rightarrow 1+\cos ^2 \gamma=1$
$\Rightarrow \cos ^2 \gamma=0$
$\Rightarrow \cos \gamma=0$
$\Rightarrow \gamma=\frac{\pi}{2}=90^{\circ}$
View full question & answer→Question 481 Mark
The equation of a line is given by $\frac{4-x}{2}=\frac{y+3}{3}=\frac{z+2}{6}$, the direction cosines of line parallel to the given line is
Answer$(d) :$ Equation of given line is $\frac{4-x}{2}=\frac{y+3}{3}=\frac{z+2}{6}$.
The direction ratios of the given line are $-2,3,6$.
$\therefore $ The direction cosines of the given line are
$\left(\frac{-2}{\sqrt{4+9+36}}, \frac{3}{\sqrt{4+9+36}}, \frac{6}{\sqrt{4+9+36}}\right)$
$=\left(\frac{-2}{\sqrt{49}}, \frac{3}{\sqrt{49}}, \frac{6}{\sqrt{49}}\right)=\left(\frac{-2}{7}, \frac{3}{7}, \frac{6}{7}\right)$
View full question & answer→Question 491 Mark
The direction cosines of the line passing through two points $(2,1,0)$ and $(1,-2,3)$ are
Answer(c) : Here, $P(2,1,0)$ and $Q(1,-2,3)$
So, $P Q=\sqrt{(1-2)^2+(-2-1)^2+(3-0)^2}$ $=\sqrt{1+9+9}=\sqrt{19}$
Thus, the direction cosines of the line joining two points are $\left\langle\frac{1-2}{\sqrt{19}}, \frac{-2-1}{\sqrt{19}}, \frac{3-0}{\sqrt{19}}\right\rangle=\left\langle\frac{-1}{\sqrt{19}}, \frac{-3}{\sqrt{19}}, \frac{3}{\sqrt{19}}\right\rangle$
View full question & answer→Question 501 Mark
Find the direction cosines of the line that makes equal angles with the three axes in space.
Answer(c) : Since $l=m=n$ and $l^2+m^2+n^2=1$
$\Rightarrow l=m=n= \pm \frac{1}{\sqrt{3}}$
View full question & answer→