Question 511 Mark
$P$ is a point on the line segment joining the points $(3,2,-1)$ and $(6,2,-2)$. If $x$ co-ordinate of $P$ is $5$ , then its $y$ co$-$ordinate is
AnswerEquation of line joining the points $(3,2,-1)$ and $(6,2,-2)$ is,
$\frac{x-3}{6-3}=\frac{y-2}{2-2}=\frac{z+1}{-2+1} \text { i.e., } \frac{x-3}{3}=\frac{y-2}{0}=\frac{z+1}{-1}=\lambda \text { (say) }$
$\Rightarrow x=3 \lambda+3, y=2, z=-\lambda-1$
So, $y$-coordinate of $P$ is 2 .
View full question & answer→Question 521 Mark
Find the equation of the plane passing through the points P(1, 1, 1), Q(3, -1, 2), R(-3, 5, -4):
Question 531 Mark
Find the shortest distance between the given two lines :
$\frac{x+1}{1}=\frac{y+1}{-1}=\frac{z+1}{1}$ and $\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-4}{4}$.
Answer(a) : $x_1=-1, y_1=-1, z_1=-1, a_1=1, b_1=-1, c_1=1$ and $x_2=2, y_2=3, z_2=4, a_2=2, b_2=3, c_2=4$
$
\begin{aligned}
\therefore \quad & d=\left|\frac{\left|\begin{array}{ccc}
3 & 4 & 5 \\
1 & -1 & 1 \\
2 & 3 & 4
\end{array}\right|}{\sqrt{(-4-3)^2+(2-4)^2+(3+2)^2}}\right| \\
& =\left|\frac{3(-4-3)-4(4-2)+5(3+2)}{\sqrt{49+4+25}}\right| \\
& =\left|\frac{-21-8+25}{\sqrt{78}}\right|=\frac{4}{\sqrt{78}} \text { units }
\end{aligned}
$
View full question & answer→Question 541 Mark
Find the direction cosines of the line joining $A(0,7,10)$ and $B(-1,6,6)$.
Answer(a) : Direction ratios of $A B$ are
$(-1-0,6-7,6-10)$ or $(-1,-1,-4)$
Also, $\sqrt{(-1)^2+(-1)^2+(-4)^2}=3 \sqrt{2}$
$\therefore \quad$ Direction cosines are $\left(-\frac{1}{3 \sqrt{2}},-\frac{1}{3 \sqrt{2}}, \frac{-4}{3 \sqrt{2}}\right)$
or $\left(\frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}, \frac{4}{3 \sqrt{2}}\right)$
View full question & answer→Question 551 Mark
Choose the correct answer from the given four options.
The locus represented by xy + yz = 0 is:
Answer
- A pair of perpendicular planes.
Solution:
We have, xy + yz = 0
⇒ xy = -yz
So, a pair of perpendicular planes.
View full question & answer→Question 561 Mark
The equation of the plane parallel to the lines x - 1 = 2y - 5 = 2z and 3x = 4y - 11 = 3z -4 and passing through the point (2, 3, 3) is:
Answer
- x - 4y + 2z + 4 = 0
Solution:
Let a, b, c be the dirction ratios of the required plane.
The given line equation can be rewritten as
$\frac{\text{x}-1}{1}=\frac{\text{y}-\frac{5}{2}}{\frac{1}{2}}=\frac{\text{z}-0}{\frac{1}{2}}\ .....(1)$
$\frac{\text{x}-0}{\frac{1}{3}}=\frac{\text{y}-\frac{11}{4}}{\frac{1}{4}}=\frac{\text{z}-\frac{4}{3}}{\frac{1}{3}}\ .....(2)$
Since the required plane is parallel to the lines (1) and (2),
$\text{a}+\frac{\text{b}}{2}+\frac{\text{c}}{2}=0\Rightarrow2\text{a}+\text{b}+\text{c}=0....(3)$
$\frac{\text{a}}{1}=\frac{\text{b}}{-4}=\frac{\text{c}}{2}=0\Rightarrow4\text{a}+3\text{b}+4\text{c}=0....(4)$
Solving (3) and (4) using cross-multiplication method, we get
$\frac{\text{a}}{1}=\frac{\text{b}}{-4}=\frac{\text{c}}{2}=\lambda\text{(say)}$
$\Rightarrow\text{a}=\lambda,\text{b}=-4\lambda,\text{c}=2\lambda$
Now, the eqution of the plane whose direction ratios are $\lambda,-4\lambda,2\lambda$ and passing through the point.
$\lambda(\text{x}-2)+(-4\lambda)(\text{y}-3)+2\lambda(\text{z}-3)=0$
$\Rightarrow\text{x}-4\text{y}+2\text{z}+4=0$ View full question & answer→Question 571 Mark
If O is the origin, OP = 3 with direction ratios proportional to -1, 2, -2 then the coordinates of P are:
Answer
- $(-1, 2,-2)$
Solution:
Let the coordinates of P be (x, y, z). Then,
Direction ratios of OP = Coordinates of P-Coordinates of O-1, 2, 2 = (x - 0), (y - 0), (z - 0)
Thus, coordinates of P are (-1, 2, -2). View full question & answer→Question 581 Mark
The distance of the point P(a, b, c) from the x-axis is:
Answer
- $\sqrt{\text{b}^2+\text{c}^2}$
Solution:
The projection of the point P(a, b, c) on the x-axis is a, (0, 0) as both Y and Z coordinates on any point on the x-axis are equal to zero.
$\therefore$ Distance of P(a, b, c) from x-axis = Distance of P(a, b, c) from a, (0, 0)
$=\sqrt{(\text{a}-\text{a})^2+(\text{b}-0)^2+(\text{c}-0)^2}$
$=\sqrt{\text{b}^2+\text{c}^2}$ View full question & answer→Question 591 Mark
The vector equation of the line passing through the point (-1, 5, 4) and perpendicular to the plane z = 0 is:
Answer
- $\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
Solution:
Given,
a = (-1, 5, 4)
b = (0, 0, 1) [$\therefore$ 1 to plone z]
We know that,
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\vec{\text{r}}=(-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})+\lambda\hat{\text{k}}$
$\Rightarrow\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$ View full question & answer→Question 601 Mark
Choose the correct answer from the given four options.
If the directions cosines of a line are $k, k, k,$ then:
AnswerSince, direction cosines of a line are $k, k,$ and $k.$
$\therefore l = k, m = k$ and $n = k$
We know that, $l^2 + m^2 + n^2 = 1$
$\Rightarrow k^2 + k^2 + k^2 = 1$
$\text{k}^2=\frac{1}{3}$
$\therefore\text{k}=\pm\frac{1}{\sqrt{3}}$
View full question & answer→Question 611 Mark
The line x = 1, y = 2 is:
Question 621 Mark
A normal to the plane x = 2 is:
Answer
- < br > (0, 1, 1) < br >
Solution:
The plane x = 2 is perpendicular to x axis So the angle is $\frac{\pi}{2},\cos\frac{\pi}{2}=0$
0 The plane x = 2 is parallel to both y axis and z axis So the angle is (0, 1, 1) View full question & answer→Question 631 Mark
The distance of the line $\vec{\text{r}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}+\lambda(\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})$ from the plane $\vec{\text{r}}.(\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}})=5$ is:
Answer
- $\frac{10}{3\sqrt{3}}$
Solution:
The given line passes through the point whose position vector is $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
We know that the perpendicular distance of a point P of position vector $\vec{\text{a}}$ from the plane $\vec{\text{r}}.\vec{\text{n}}=\text{d}$ is given by
$\text{P}=\frac{\big|\vec{\text{a}}.\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
Here, $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{n}}=\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}},\text{d}=5$
So, the required distance P is given by
$\text{P}=\frac{\Big|\big(2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big),\big(\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}}\big)-5\Big|}{\Big|\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}}\Big|}$
$=\frac{|2-10+3-5|}{\sqrt{1+25+1}}$
$=\frac{|-10|}{\sqrt{27}}$
$=\frac{10}{3\sqrt{3}}\text{units}$ View full question & answer→Question 641 Mark
Which of the following triplets give the direction cosines of a line:
AnswerIf $l, m, n$ are the directions cosine of a line then $l^2+ m^2 + n^2 = 1 $
Thus we get $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
View full question & answer→Question 651 Mark
The projections of a line segment on x, y and z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are:
Answer
- $13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$
Solution:
If a line makes angles $\alpha,\beta$ and $\gamma$ with the axes, then $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\dots(1)$
Let r be the length of the line segment. then,
$\text{r}\cos\alpha=12,\text{r}\cos\beta=4,+\cos\gamma=3\dots(2)$
$\Rightarrow\big(\text{r}\cos\alpha\big)^2+\big(\text{r}\cos\beta\big)^2+\big(\text{r}\cos\gamma\big)^2=12^2+4^3+3^2$
$\Rightarrow\text{r}^2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)=169$
$\Rightarrow\text{r}^2(1)=169$ [From (1)]
$\Rightarrow\text{r}=\sqrt{169}$
$\Rightarrow\text{r}=\pm13$
$\Rightarrow\text{r}=13$ (Since length cannot be negative)
(Since legth cannot be negative)
Substituting r = 13 in (2), we get
$\cos\alpha=\frac{12}{13},\cos\beta\frac{4}{13},\cos\gamma=\frac{1}{13}$ Thus, the direction cosines of the line are $\frac{12}{13},\frac{4}{13},\frac{1}{13}$ View full question & answer→Question 661 Mark
The eqution of the plane $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\lambda(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})+\mu(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})$ in scalar product from is:
Answer
- $\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}})=7$
Solution:
We know that the equation $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ represents a plane passing through a point whose position vectors is $\vec{\text{a}}$ and parrallel to the vectors $\vec{\text{b}}$ and $\vec{\text{c}}$.
Here, $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}};\ \vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}};\ \vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&1\\1&-2&3\end{vmatrix}$
$=5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
The vector equation of the plane in scalar product from is
$\vec{\text{r}}.\vec{\text{n}}=\vec{\text{a}}.\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=\big(\hat{\text{i}}-\hat{\text{j}}-0\hat{\text{k}}\big).\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=5+2+0$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=7$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=7$ View full question & answer→Question 671 Mark
The straigth line $\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$ is:
Answer
- perpendicular to z-axis
Solution:
We have
$\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$
Also, the given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
Let $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ be parpendicular 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 681 Mark
The sum of the squares of sine of the angles made by the line AB with OX, OY, OZ where O is the origin is:
Question 691 Mark
The points (k − 1, k + 2), (k, k + 1), (k + 1, k) are collinear for:
Question 701 Mark
If a line has the direction ratio 18, 12, 4, then its direction cosines are:
Answer
- $\frac{9}{11},\frac{6}{11},\frac{2}{11}$
Solution:
Dr's of the line are : 18, 12, 4
$\text{Dc's}=\frac{18}{\sqrt{18^2+12^2+4^2}},\frac{12}{\sqrt{18^2+12^2+4^2}},\frac{4}{\sqrt{18^2+12^2+4^2}}$
$=\frac{18}{22},\frac{12}{22},\frac{4}{22}$
$=\frac{9}{11},\frac{6}{11},\frac{2}{11}$ View full question & answer→Question 711 Mark
Cosine of the angle between two diagonals of acube is equal to:
Question 721 Mark
The straight line $\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$ is:
View full question & answer→Question 731 Mark
Distance of the point $(\alpha, \beta, \gamma)$ from y-axis is:
Answer
- $\sqrt{\alpha^2+\gamma^2}$
View full question & answer→Question 741 Mark
What are the DR's of vector parallel to (2, −1, 1) and (3, 4, −1)?
Answer
- (1, 5, −2)
Solution:
Required DR's are (3 − 2, 4 + 1, −1−1) ie, (1, 5, −2)
View full question & answer→Question 751 Mark
The distance of the plane 2x - 3y + 6z + 7 = 0 from the point (2, -3, -1) is:
Question 761 Mark
The equation of the plane through point (1, 2, -3) which is parallel to the plane 3x - 5y + 2z = 11 is given by:
Question 771 Mark
The direction cosines of a line which is equally inclined to axes, is given by:
Answer
- $\underline{+}\frac{1}{\sqrt{3}}$
View full question & answer→Question 781 Mark
The three points ABC have position vectors (1, x, 3), (3, 4, 7) and (y, -2, -5) are collinear then (x, y):
Question 791 Mark
The equation of the plane passing through (2, −3, 1) and is normal to the line joining the points (3, 4, −1) and (2, −1, 5) is given by:
Question 801 Mark
If P be the point (2, 6, 3) then the equation of the plane trough P, at right angles to OP, where ′O′ is the origin is:
Question 811 Mark
If a line makes 45°, 60° with positive direction of axes x and y then the angles it makes with the z-axis is:
Question 821 Mark
The coordinates of the midpoints of the line segment joining the points (2, 3, 4) and (8, -3, 8) are:
Question 831 Mark
If $\alpha, \beta, \gamma$ are the direction angles of a vector and $\cos \alpha=\frac{14}{15}, \cos \beta=\frac{1}{3}$, then $\cos \gamma=$
Answer$\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
$(\because \alpha, \beta, \gamma$ are direction angles $)$
$\Rightarrow \frac{196}{225}+\frac{1}{9}+\cos ^2 \gamma=1$
$\Rightarrow \cos ^2 \gamma=1-\frac{221}{225}=\frac{4}{225}$
$\Rightarrow \cos \gamma= \pm \frac{2}{15}$
View full question & answer→Question 841 Mark
The vector equation of the line through the points $A(3,4,-7)$ and $B(1,-1,6)$ is
Answer(c): If $\vec{a}=P . V$. of $A=3 \hat{i}+4 \hat{j}-7 \hat{k}$
and $\vec{b}=$ P.V. of $B=\hat{i}-\hat{j}+6 \hat{k}$, then the equation of line $A B$ is $\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$
$
\therefore \quad \vec{r}=(3 \hat{i}+4 \hat{j}-7 \hat{k})+\lambda(-2 \hat{i}-5 \hat{j}+13 \hat{k})
$
View full question & answer→Question 851 Mark
The plane $2\text{x}-(1-\lambda)\text{y}+3\lambda\text{z}=0$ passes through the intersection of the planes:
Answer
- 2x - y = 0 and y- 3z = 0
Solution:
The given plane is
$2\text{x}-(1-\lambda)\text{y}+3\lambda\text{z}=0$
$\Rightarrow(2\text{x}-\text{y})+\lambda(-\text{y}+3\text{z})=0$
So, this plane passes through the intersection of the planes
2x - y = 0 and -y + 3z = 0
⇒ 2x - y = 0 and y - 3z = 0. View full question & answer→Question 861 Mark
Are the points $(1, 1), (2, 3)$ and $(8, 11)$ collinear$?$
AnswerArea of triangle formed by these vertices is,
$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\2&3&1\\8&11&1\end{vmatrix}$
Applying $R_2 \rightarrow R_2 − R_1, R_{3 }\rightarrow R_{3 }− R_1$
$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\1&2&0\\7&10&0\end{vmatrix}=\frac{1}{2}(10-14)=2$
Hence points are non collinear
View full question & answer→Question 871 Mark
If P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10) are collinear, then R divided PQ in the ratio:
Answer
- 3 : 2 externally
Solution:
Suppose the point R divides PQ in the ratio $\lambda:1$.
Coordinates of R are $\Big(\frac{5\lambda+3}{\lambda+1},\frac{4\lambda+2}{\lambda+1},\frac{-6\lambda-4}{\lambda+1}\Big)$.
But the coordinates of R are (9, 8, -10).
$\therefore\frac{5\lambda+3}{\lambda+1}=9,\frac{4\lambda+2}{\lambda+1}=8$ and $\frac{-6\lambda-4}{\lambda+1}=-10$
From each of these equations, we get
$\lambda=-\frac{3}{2}$
$\therefore$ R divided PQ in the ratio 3 : 2 externally. View full question & answer→Question 881 Mark
Choose the correct answer from the given four options.
The reflection of the point $(\alpha,\beta,\gamma)$ in the xy–plane is:
Answer
- $(\alpha,\beta,-\gamma)$
Solution:
In XY-plane, only the sign of z coordinate of the point got changed after the reflection. Therefore, the reflection of the point $(\alpha,\beta,\gamma)$ is $(\alpha,\beta,-\gamma).$ View full question & answer→Question 891 Mark
If the lines $\text{ x - }\frac{2}{1} =\text{y}-\frac{2}{1} =\text{z}-\frac{4}{\text{k}} $ and $\text{x}-\frac{1}{\text{k}} = \text{y}-\frac{4}{2} = \text{z}-\frac{5}{1} $ are coplanar, then k can have:
View full question & answer→Question 901 Mark
The direction ratios of the line 6x - 2 = 3y + 1 = 2z - 2 are:
Answer
- $1, 2, 3$
Solution:
6x - 2 = 3y + 1 = 2x - 2
$6\big(\text{x}-\frac{2}{6}\big)=3(\text{y}+\frac{1}{3}\big)=2(\text{x}-\frac{2}{2}\big)$
$\frac{\big(\text{x}-\frac{1}{3}\big)}{1}=\frac{\big(\text{y}+\frac{1}{3}\big)}{2}=\frac{(\text{x}-1)}{3}$
Line will be passing through the poits, $\Big(\frac{1}{3},-\frac{1}{3},1\Big)$
and parallel to the line having direction ratios is 1, 2, 3 View full question & answer→Question 911 Mark
If the three points A(1, 6), B(3, −4) and C(x, y) are collinear, then the equation satisfying by x and y is:
Answer
- 5x + y − 11 = 0
Solution:
Since, the points A(1, 6), B(3, −4) and C(x, y) are colinear
$\therefore\begin{vmatrix}1&6&1\\3&-4&1\\\text{x}&\text{y}&1\end{vmatrix}=0$
⇒ 1(−4−y) −6(3 − x) + 1(3y + 4x) = 0
⇒ 10x + 2y − 22 = 0
⇒ 5x + y − 11 = 0 View full question & answer→Question 921 Mark
The equation of the line passing through the points $\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$ and $\text{b}\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ is:
Answer
- $\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
Solution:
Equation of the line passing through the points having position vectors
$\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$ and $\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ is:
$\vec{r}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big\{\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)-\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)\big\},$ where t is a parameter
$=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)-\text{t}\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
$=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$ View full question & answer→Question 931 Mark
A parallelopiped is formed by planes drawn through the point (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes. The length of a diagonal of the parallelopiped is:
Answer
- 7
Solution:
The given point (2, 3, 5) and (5, 9, 7) are two diagonally opposite vertices of the parallelopiped as all of theire coordinates are different.
$\therefore$ Edges of the paralleloppiped
= |2 - 5|, |3 - 9| and |5 - 7|
=3, 6 and 2.
Now,
Length of the diagonal of the parallelopiped
$=\sqrt{3^2+6^2+2^2}$
$=\sqrt{9+36+4}$
$=\sqrt{49}$
$=7$
Hence, length of the diagonal of the parallelepiped formed by the planes
Parallel to coordinate planes and drawn through point (2, 3, 5)and (5, 9, 7) is 7 units. View full question & answer→Question 941 Mark
The vector equation of the plane containing the line $\vec{\text{r}}=(-2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})+\lambda(3\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$ and the point $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ is:
Answer
- $\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$
Solution:
Let the direction ratio of the required plane be proportinal to a, b, c.
Scince the required plane contains the line $\vec{\text{r}}=(-2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})+\lambda(3\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
It must pass through the point (-2, -3, 4) and it should be parallel to the line.
So, the equation of the plane is
a(x + 2) + b(y + 3) + c(z - 4) = 0 ....(1) and
3a - 2b - c = 0 ....(2)
It is given that plane (1) passes through the point $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ or (1, 2, 3).
a(1 + 2) + b(2 + 3) + c(3 - 4) = 0
3a + 5b - c = 0 .......(3)
So,
Solving (1) (2) and (3), we get
$\begin{vmatrix}\text{x}+2&\text{y}+3&\text{z}-4\\3&-2&-1\\3&5&-1\end{vmatrix}=0$
$\Rightarrow7(\text{x}+2)+0(\text{y}+3)+21(\text{y}-4)=0$
$\Rightarrow\text{x}+2+3\text{z}-12=0$
$\Rightarrow\text{x}+3\text{z}=10$ or $\vec{\text{r}}.\big(\hat{\text{i}}+3\hat{\text{k}}\big)=10$ View full question & answer→Question 951 Mark
The eqution of the plane through the line $x + y + 3 = 0 = 2x - y + 3z + 1$ and parallel to the line $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ is:
AnswerThe equation of the plane passing though the line of intersection of the given planes is
$\text{x}+\text{y}+\text{z}+3+\lambda(2\text{x}-\text{y}+3\text{z}+1)=0$
$(1+2\lambda)\text{x}+(1-\lambda)\text{y}+(1+3\lambda)\text{z}+3+\lambda=0\ ....(1)$
This plane is parallel to the line $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}.$
It means that this line is perpendicular to the normal of the plane $(1).$
$\Rightarrow1(1+2\lambda)\text{x}+2(1-\lambda)+3(1+3\lambda)=0($Because $a_1a_2 + b_1b_2 + c_1c_2 = 0)$
$\Rightarrow1+2\lambda+2-2\lambda+3+9\lambda=0$
$\Rightarrow9\lambda+6=0$
$\Rightarrow\lambda=\frac{-2}{3}$
Substituting this in $(1),$ we get
$\Big(1+2\Big(\frac{-2}{3}\Big)\Big)\text{x}+\Big(1-\Big(\frac{-2}{3}\Big)\Big)\text{y}+\Big(1+3\Big(\frac{-2}{3}\Big)\Big)\text{z}+3+\Big(\frac{-2}{3}\Big)=0$
$\Rightarrow -x + 5y - 3z + 7 = 0$
$\Rightarrow x - 5y + 3z = 7$
View full question & answer→Question 961 Mark
A vector parallel to the line of intersection of the plance $\vec{\text{r}}.(3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=1$ and $\vec{\text{r}}.(\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}})=2$ is:
Answer
- $2\hat{\text{i}}+7\hat{\text{j}}-13\hat{\text{k}}$
Solution:
Let the required vector be a $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\ ....(1)$
Since the vector is parallel to the line of intersection of the given planes,
3a - b + c = 0 .....(2)
a + 4b - 2c = 0 ....(3)
Solving (2) and (3), we get
$\frac{\text{a}}{-2}=\frac{\text{b}}{7}=\frac{\text{c}}{13}$
Substituting these values in (1), we get
$-2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$ which is the required vector. View full question & answer→Question 971 Mark
The points A(1, 1, 0), B(0, 1, 1), C(1, 0, 1) and $\text{D}\big(\frac{2}{3},\frac{2}{3},\frac{2}{3}\big)$
View full question & answer→Question 981 Mark
The equation of the plane through the origin and parallel to the plane 3x - 4y + 5z + 6 = 0:
Question 991 Mark
The distance of the plane through the intersection of the planes ax + by + cz +d = 0 and lx + my + nz + P = 0 and parallel to the line y = 0, z = 0
Answer
- (bl - am)y + (cl - an)z + dl - ap = 0
Solution:
The equation of the plane passing through the intersection of the planes
ax + by + cz + d = 0
and lx + my + nz + p =0
Will be $(\text{ax} + \text{by} +\text{cz} +\text{d})+\lambda(\text{lm}+\text{my}+\text{nz}+\text{p})=0$
$\text{x}(\text{a}+\lambda1)+\text{y}(\text{b}+\lambda\text{m})+\text{z}(\text{c}+\lambda\text{n})+(\text{d}+\lambda\text{p})=0\ (1)$
Since the plane is parallel to the line y = 0 and z = 0
$\text{a}+\lambda1=0$
$\lambda=\frac{-\text{a}}{\text{l}}$
Putting the value of A in eqution (1), we get
$\text{x}\Big(\text{a}+\Big(\frac{\text{-a}}{\text{l}}\Big)\text{l}\Big)+\text{y}\Big(\text{b}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{m}+\text{y}\Big(\text{c}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{n}+\text{d}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{p}=0$
$\text{y}(\text{bl}-\text{am})+\text{z}(\text{cl}-\text{an})+\text{dl}-\text{ap}=0$
Heance, option (a) View full question & answer→Question 1001 Mark
The eqution of the plane contaning the two lines $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}-0}{3}$ and $\frac{\text{x}}{-2}=\frac{\text{y}-2}{-3}=\frac{\text{z}+1}{-1}$ is:
Answer
- None of these
Solution:
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}-0}{3}$ and $\frac{\text{x}}{-2}=\frac{\text{y}-2}{-3}=\frac{\text{z}+1}{-1}$
Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector. View full question & answer→