Question 1011 Mark
If a line makes angles $Q_1, Q_{21}$ and $Q_3$ respectively with the coordinate axis then the value of $\cos^2 \text{Q}_{1} + \cos^2 \text{Q}_{2} + \cos^2 \text{Q}_{3}$:
View full question & answer→Question 1021 Mark
The reflection of the point $(\text{a}, \beta, \gamma) $ in the xy-plane is:
View full question & answer→Question 1031 Mark
A line OP where O = (0, 0, 0) makes equal angles with ox, oy, oz. The point on OP, which is at a distance of 6 units from O is:
Answer
- $\Big(\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}}\Big)$
View full question & answer→Question 1041 Mark
ox, oy are positive x-axis, positive y-axis respectively where O = (0, 0,0) The d.c.s of the llne which bisects $\angle\text{xoy}$ are:
Answer
- $\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0$
Solution:
Equation of line bisecting XOY is x = y
$\therefore$ d.r.s are (1, 1, 0)
And thus d.c.s are $\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$ View full question & answer→Question 1051 Mark
What is the sum of the squares ofdirection cosines of the line joining thepoints (1, 2, -3) and (-2, 3, 1):
Answer
- 1
Solution:
The sum of the squares of direction cosines of the line is always 1
View full question & answer→Question 1061 Mark
If $\alpha,\beta$ and $\gamma$ are the angles which a half ray makes with the positive direction of the axes, then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ is equal to:
Answer
- 2
Solution:
Given expression, $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$
$=(1-\cos^2\alpha)+(1-\cos^2\beta)+(1-\cos^2\gamma)$
$=3-\cos^2\alpha+\cos^2\beta+\cos^2\gamma=3-1=2$
$(\because\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1)$ View full question & answer→Question 1071 Mark
The direction ratios of the line x - y + z - 5 = 0 = x - 3y - 6 are proportional to:
Answer
- $3,1,-2$
Solution:
We have
x - y + z - 5 = 0 = x - 3y - 6
⇒ x - 3y - 6=0
x - y + z - 5 = 0
⇒ x = 3y + 6 .....(1)
x - y + z - 5 = 0.....(2)
From (1) and (2), we get
3y + 6 - y + z - 5 = 0
⇒ 2y + z + 1 = 0
$\Rightarrow\text{y}=\frac{-\text{z}-1}{2}$
$\text{y}=\frac{\text{x}-6}{3}$ [From (1)]
$\therefore\frac{\text{x}-6}{3}=\text{y}=\frac{-\text{z}-1}{2}$
So, the given equation can be re-witten as
$\frac{\text{x}-6}{3}=\frac{\text{y}}{1}=\frac{\text{z}+1}{-2}$
Hence, the direction ratios the given line are proportional to 3, 1, -2. View full question & answer→Question 1081 Mark
If $\alpha,\beta,\gamma$ are the angles which a directed line makes with the positive directions of the coordinate axes, then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ is equal to:
Answer
- 2
Solution
The direction cosines of the line are
$\text{l}^2+\text{m}^2+\text{n}^2=1$
Now, $\Rightarrow\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$
$\Rightarrow1-\sin^2\alpha+1-\sin^2\beta+1-\sin^2\gamma=1$
$\Rightarrow\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2$ View full question & answer→Question 1091 Mark
If a line makes angle $\frac{\pi}{3}$ and $\frac{\pi}{4}$ with x-axis and y-axis respectively, then the angle made by the line with z-axis is:
Answer
- $\frac{\pi}{3}$
Solution:
If a line makes angles $\alpha,\beta$ and $\gamma$ with the axcs, then $\cos2\alpha+\cos2\beta+\cos2\gamma=1.$
Here,
$\alpha=\frac{\pi}{3}$
$\beta=\frac{\pi}{4}$
Now,
$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$
$\Rightarrow\cos^2\frac{\pi}{3}+\cos^2\frac{\pi}{4}+\cos^2\gamma=1$
$\Rightarrow\frac{1}{4}+\frac{1}{2}+\cos^2\gamma=1 $
$\Rightarrow\cos^2\gamma=1-\frac{3}{4}$
$\Rightarrow\cos^2\gamma=\frac{1}{4}$
$\Rightarrow\cos\gamma=\frac{1}{2}$
$\Rightarrow\gamma=\frac{\pi}{3}$ View full question & answer→Question 1101 Mark
The direction ratios of two lines AB, AC are 1, -1, -1 and 2, -1, 1. The direction ratios of the normal to the plane ABC are:
Question 1111 Mark
(2, -3, -1) 2x - 3y + 6z + 7 = 0:
Question 1121 Mark
If $\cos\alpha,\cos\beta,\cos\gamma$ are the direction cosines of a vector $\vec{\text{a}}$ then $\cos2\alpha+\cos2\beta+\cos2\gamma$ is equal to:
View full question & answer→Question 1131 Mark
If the projections of the line segment $AB$ on the coordinate axes are $12, 3, k$ and $AB = 13$ then $k^2 - 2k + 3$ is equal to:
AnswerLet $a, b, c$ be the projection of a line on the coordinate axes.
Then the length of the line given by $\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}$
Here we have $12^2 + 3^2 + k^2 = 169$
$\Rightarrow\text{k}=\underline{+}4$
Thus $k^2 - 2k + 3 = 11$ or $27.$
View full question & answer→Question 1141 Mark
The distance between the planes $2x + 2y - z +2 = 0$ and $4x + 4y - 2z + 5 = 0$ is:
AnswerMultiplying the first equation of the plane by
$4x + 4y - 2z + 4 = 0$
$4x + 4y - 2z = -4 .....(1)$
The second eqution of the plane is
$4x + 4y - 2z + 5 = 0$
$4x + 4y - 2z = -5 .....(2)$
We know that the distance between two planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is,
$=\frac{|\text{d}_2-\text{d}_1|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|-5+4|}{\sqrt{4^2+4^2+(-2)^2}}$
$=\frac{|-1|}{\sqrt{16+16+4}}$
$=\frac{1}{\sqrt{36}}$
$=\frac{1}{6}\text{units}$
View full question & answer→Question 1151 Mark
lf a line makes angles $\frac{\pi}{12},\frac{5\pi}{12}$ with oy, oz respectively where 0 = (0, 0, 0), then the angle made by that line with ox is:
Answer
- 90°
Solution:
$\Big(\cos\frac{\pi}{12}\Big)^2+\Big(\cos\frac{5\pi}{12}\Big)^2+\big(\cos(\gamma)\big)^2=1$
$\Big(\cos\frac{\pi}{12}\Big)^2+\Big(\cos\frac{\pi}{12}\Big)^2+\big(\cos(\gamma)\big)^2=1..$
$\Big(\cos\theta=\sin\Big(\frac{\pi}{2}- \theta\Big)\Big)$
$\Big(\cos(\gamma)\Big)^2=0$
$\cos(\gamma)=0$
$\gamma=90^\circ$ View full question & answer→Question 1161 Mark
If a line makes the angle $\alpha,\beta,\gamma$ with three dimensional coordinate axes respectively, then $\cos2\alpha+\cos2\beta+\cos2\gamma$ is equal to:
Answer
- -1
Solution:
We need to find value of $\cos2\alpha+\cos2\beta+\cos2\gamma$
It is further equal to $\cos^2\alpha-1+\cos^2\beta-1+\cos^2\gamma-1$
$=2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)-3$
$= 2(1) - 3 = 2 = -1$
$\therefore(\text{l}^2 + \text{m}^2 + \text{n}^2 = 1)$ View full question & answer→Question 1171 Mark
lf $\text{AB}\perp\text{BC}$ then the value of $\lambda$ equal, where A(2k, 2, 3), B(k, 1, 5), C(3 + k, 2, 1):
Answer
- $-3$
Solution:
The drs of AB are (k, 1, -2)
The drs of BC are (3, 1, -4)
Since, they are perpendicular, AB.BC = 0
3k + 1 + 8 = 0
k = -3 View full question & answer→Question 1181 Mark
The projection of the join of the two points (1, 4, 5), (6, 7, 2) on the line whose d.ss are (4, 5, 6) is:
Question 1191 Mark
If the direction ratios of two lines are given by 3lm - 4ln + mn = 0 and l + 2m + 3n = 0, then the angle between the lines is:
Question 1201 Mark
The vector equation of the line $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is
Answer(a): We have, $\vec{a}=5 \hat{i}-4 \hat{j}+6 \hat{k}$ and $\vec{b}=3 \hat{i}+7 \hat{j}+2 \hat{k}$
Therefore, the vector equation will be $\vec{r}=\vec{a}+\lambda \vec{b}$
$
\Rightarrow \vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})
$
View full question & answer→Question 1211 Mark
The perpendicular distance of the point P(1, 2, 3) from the line $\frac{\text{x}-6}{3}=\frac{\text{y}-7}{2}=\frac{\text{z}-7}{-2}$ is:
Answer
- 7
Solution:
$\frac{\text{x}-6}{3}=\frac{\text{y}-7}{2}=\frac{\text{z}-7}{-2}$
Let point (1, 2, 3) be P and the point through which the line passes be Q(6, 7, 7). Also, the line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$
Now,
$\overrightarrow{\text{PQ}}=5\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
$\therefore\vec{\text{b}}\times\overrightarrow{\text{PQ}} =\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&2&-2\\5&5&4\end{vmatrix}$
$=18\hat{\text{i}}-22\hat{\text{j}}+5\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|=\sqrt{18^2+(-22)^2+5^2}$
$=\sqrt{324+484+25}$
$=\sqrt{833}$
$\because\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}$
$=\frac{\sqrt{833}}{\sqrt{17}}$
$=\sqrt{49}$
$=7$ View full question & answer→Question 1221 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.
$\text{As}\Big(\frac{1}{\sqrt{3}}\Big)^2+\Big(\frac{2}{\sqrt{3}}\Big)^2+\Big(\frac{-2}{\sqrt{3}}\Big)^2$
$=\frac{1+4+4}{3}$
$=3$ View full question & answer→Question 1231 Mark
The shortest distance between the lines $\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}$ and, $\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}$ is:
Answer
- $3\sqrt{30}$
Solution:
We have
$\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}\dots(1)$
$\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}\dots(2)$
We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3, -1, 1.
Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1,$ where $\vec{\text{a}}_1=3\hat{\text{i}}+8\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{b}}_1=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}.$
Also, line (2) passes through the point (3, -7, 6) and has direction ratios proprtional to -3, 2, 4.
Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2,$ where $\vec{\text{a}}_2=-3\hat{\text{i}}-7\hat{\text{j}}+6\hat{\text{k}}$ and $\vec{\text{b}}_2=-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}.$
Now,
$\vec{\text{a}}_2-\vec{\text{a}}_1=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-1&1\\-3&2&4\end{vmatrix}$
$=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-6)^2+(-15)^2+3^2}$
$=\sqrt{36+225+9}$
$=\sqrt{270}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big).\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big)$
$=36+225+9$
$=270$
The shortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{270}{\sqrt{270}}\Big|$
$=\sqrt{270}$
$=3\sqrt{30}$ View full question & answer→Question 1241 Mark
The direction cosines of the y-axis are:
Question 1251 Mark
The projection of a directed line segment on the co-ordinate axes are 12, 4, 3, then the direction cosines of the line are:
Answer
- $\frac{12}{13},\frac{4}{13},\frac{3}{13}$
Solution:
x = 12, y = 4, z = 3
Direction cosines =
$\frac{\text{x}}{\text{x}^2 + \text{y}^2+\text{z}^2},\frac{\text{y}}{\text{x}^2 + \text{y}^2+\text{z}^2},\frac{\text{x}}{\text{x}^2 + \text{y}^2+\text{z}^2}$
$=\frac{12}{13},\frac{4}{13},\frac{3}{13}$ View full question & answer→Question 1261 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 $\frac{x-0}{1 / 2}=\frac{y-0}{1 / 3}=\frac{z-0}{-1}$ and $\frac{x-0}{1 / 6}=\frac{y-0}{-1}=\frac{z-0}{-1 / 4}$
$\therefore a_1=\frac{1}{2}, b_1=\frac{1}{3}, c_1=-1$
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 1271 Mark
Find the equation of a line passing through $(1,2,-3)$ and parallel to the line $\frac{x-2}{1}=\frac{y+1}{3}=\frac{z-1}{4}$.
Answer(b) : Since, the line is parallel to the line
$
\frac{x-2}{1}=\frac{y+1}{3}=\frac{z-1}{4} \text {. }
$
$\therefore \quad$ D.r.'s of the required line are < 1,3,4 >.
Hence, equation of the line passing through $(1,2,-3)$ with d.r.'s < 1,3,4 > is $\frac{x-1}{1}=\frac{y-2}{3}=\frac{z+3}{4}$
View full question & answer→Question 1281 Mark
The cosines of the angle between any two diagonals of a cube is:
Question 1291 Mark
For every point P(x, y, z) on the x-axis (except the origin),
Answer
- y = 0, z = 0, x ≠ 0
Solution:
Both Y and Z coordinates on each point of the x-axis are equal to zero.
The X-coordinate on the origin is also equal to zero.
Therefore, the Y and Z coordinates on each point of the x-axis, except the origin, are equal to zero,
While the X-coordinate is non-zero.
View full question & answer→Question 1301 Mark
The distance of the points (2, 1, -1) from the plane x - 2y + 4z - 9 is:
Question 1311 Mark
If the equation of a line $A B$ is $\frac{x-3}{1}=\frac{y+2}{-2}$ $=\frac{z-5}{4}$, find the direction ratios of a line parallel to $A B$.
Answer(d) : The direction ratios of line parallel to $A B$ is $1,-2$ and 4 .
View full question & answer→Question 1321 Mark
The direction ratios of the line joining the points $(x, y, z)$ and $(x_2, y_2, z_1)$ are:
Answer$\text{x}_{2} - \text{x}_{1}, \text{y}_{2} - \text{y}_{1}, \text{z}_{2} -\text{ z}_{1}$
View full question & answer→Question 1331 Mark
The direction ratios of the line perprndicular to the lines $\frac{\text{x}-7}{2}=\frac{\text{y}+17}{-3}=\frac{\text{z}-6}{1}$ and, $\frac{\text{x}+5}{1}=\frac{\text{y}+3}{2}=\frac{\text{z}-4}{-2}$ are proportional to:
Answer
- 4, 5, 7
Solution:
We have
$\frac{\text{x}-7}{2}=\frac{\text{y}+17}{-3}=\frac{\text{z}-6}{1}$
$\frac{\text{x}+5}{1}=\frac{\text{y}+3}{2}=\frac{\text{z}-4}{-2}$
The direction ratios of the given lines are proportional to 2, -3, 1 and 1, 2, -2.
The vectors parallel to the given vectors are $\vec{\text{b}}_1=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$
Vector perpendicular to the given two lines is
$\vec{\text{b}}=\vec{\text{b}}_1\times\vec{\text{b}}_2$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-3&1\\1&2&-2\end{vmatrix}$
$=4\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}$
Hence, the direction ration of the line perpendicular to the given two lines are proportional to 4, 5, 7. View full question & answer→Question 1341 Mark
The cartesian equation of the line which passes through the point $(-2,4,-5)$ and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ is
Answer(a) : It is given that the line passes through the point $(-2,4,-5)$ and is parallel to
$
\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6} \text {. }
$
Clearly, the direction ratios of line are $(3,5,6)$.
Now the equation of the line (in cartesian form) is
$
\frac{x-(-2)}{3}=\frac{y-4}{5}=\frac{z-(-5)}{6} \Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}
$
View full question & answer→Question 1351 Mark
The angle between the lines $\frac{\text{x}-1}{1}=\frac{\text{y}-1}{1}=\frac{\text{z}-1}{2}$ and $\frac{\text{x}-1}{-\sqrt{3}-1}=\frac{\text{y}-1}{\sqrt{3}-1}=\frac{\text{z}-1}{4}$ is:
Answer
- $\frac{\pi}{3}$
Solution:
We have
$\frac{\text{x}-1}{1}=\frac{\text{y}-1}{1}=\frac{\text{z}-1}{2}$
$\frac{\text{x}-1}{-\sqrt{3}-1}=\frac{\text{y}-1}{\sqrt{3}-1}=\frac{\text{z}-1}{4}$
The direction ratios of the given lines are proportional to 1, 1, 2 and $-\sqrt{3}-1,\sqrt{3}-1, 4$
The given lines are parallel to vectors $\vec{\text{b}}_1=\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}_2=\big(-\sqrt{3}-1\big)\hat{\text{i}}+\big(\sqrt{3}-1\big)\hat{\text{j}}+4\hat{\text{k}}$
Let $\theta$ be the angle between the given lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big).\big\{\big(-\sqrt{3}-1\big)\hat{\text{i}}+\big(\sqrt{3}-1\big)\hat{\text{j}}+4\hat{\text{k}}\big\}}{\sqrt{1^2+1^2+1^2}\sqrt{\big(-\sqrt{3}-1\big)^2+\big(\sqrt{3}-1\big)^2+4^2}}$
$=\frac{-\sqrt{3}-1+\sqrt{3}-1+8}{\sqrt{3}\sqrt{24}}$
$=\frac{6}{6\sqrt{2}}$
$\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\frac{\pi}{3}$ View full question & answer→Question 1361 Mark
If 2x + 5y - 6z + 3 = 0 be the equation of the plane, then the equation of any plane parallel to the given plane is:
Question 1371 Mark
A straight line passes through (1, -2, 3) and perpendicular to the plane 2x + 3y - z = 7. Find the direction ratios of normal to plane:
Answer
- < 2, 3, -1 >
Solution:
concept: for any plane ax + by + cz + d =
0, normal vector to this plane is $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$
the normal vector of the plane 2x + 3y - z = 7 is $\text{2}\hat{\text{i}}+\text{3}\hat{\text{j}}+\hat{\text{k}}$
so the direction ratios of normal to plane are < 2, 3, -1 > View full question & answer→Question 1381 Mark
The equation of the plane passing through the points (3, 2, −1), (3, 4, 2) and (7, 0, 6) is 5x + 3y −2z = λ where λ is:
Question 1391 Mark
If l, m, n are the d.cs of the line joining (5, -3, 8) and (6, -1, 6) then l + m + n =
Answer
- $\frac{1}{3}$
Solution:
The line joining (5, -3, 8) and (6, -1, 6) is given by the vector -i + 2j - 2k.
the direction cosines are given by. l =
$\frac{1}{\sqrt{1^2+2^2+(-2)^2}}=\frac{1}{3}, \text{m}=\frac{2}{\sqrt{1^2+2^2+(-2)^2}}=\frac{2}{3}$
$\text{n}=\frac{-2}{1^2+2^2+(-2)^2}=\frac{-2}{3}$
$\Rightarrow\text{l + m + n}=\frac{1}{3}$ View full question & answer→Question 1401 Mark
If points (1, 2), (3, 5) and (0, b) are collinear the value of b is:
Answer
- $\frac{1}{2}$
Solution:
Area $=\frac{1}{2}|1(5-\text{b})+3(\text{b}-2)+0(2-5)|$
As points are collinear, so area = 0
$\therefore\frac{1}{2}|1(5-\text{b})+3(\text{b}-2)+0(2-\text{5})|=0$
⇒ 5 − b + 3b − 6 = 0
⇒ = 1 = 2b
$\therefore\text{b}=\frac{1}{2}$ View full question & answer→Question 1411 Mark
View full question & answer→Question 1421 Mark
If $l, m, n$ are the direction cosines of a line, then:
Answer$l^{2 }+ m^{2 }+ n^2$
$ = 1$
View full question & answer→Question 1431 Mark
A straight line L on the xy-plane bisects the angle between OX and OY. What are the direction cosines of L:
Answer
- $\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$
Solution
L makes an angle $\frac{\pi}{4}$ with X and Y axis and $\frac{\pi}{2}$
$\therefore$ d.cs are $\Big(\cos\frac{\pi}{34},\cos\frac{\pi}{4},\cos\frac{\pi}{2}\Big)=\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$ View full question & answer→Question 1441 Mark
If P(x, y, z) moves such that x = 0, z = 0 then the locus of P is the line whose d.cs are:
Answer
- 0, 1, 0
Solution:
When P moves then x = 0, z = 0 but y is not given. Let y = y Then the coordinates of the point will be (0, y, 0) Now, direction cosines with respect to (0, y, 0) is given by.
$\cos\alpha=\frac{0}{0^2+\text{y}^2+0^2}=\frac{0}{\text{y}}=0$
$\cos\beta=\frac{\text{y}}{0^2+\text{y}^2+0^2}=\frac{\text{y}}{\text{y}}=1$
$\cos\gamma=\frac{{0}}{0^2+\text{y}^2+0^2}=\frac{{0}}{\text{y}}=0$
The direction cosines are 0, 1, 0 View full question & answer→Question 1451 Mark
The direction cosines of the straight linegiven by the planes x = 0 and z = 0 are:
Answer
- 0, 1, 0
Solution:
Given, x = z = 0
It represents Z-axis
$\therefore$ Direction cosines = (0, 1, 0) View full question & answer→Question 1461 Mark
The following lines are $\hat{\text{r}}=\Big(\hat{\text{i}}+\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\Big)+\mu\Big(-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\Big)$
View full question & answer→Question 1471 Mark
The direction cosines of the normal to the plane 2x - 3y - 6z - 3 = 0 are:
Answer
- $\frac{2}{7},\frac{-3}{7},\frac{-6}{7}$
View full question & answer→Question 1481 Mark
If $\left(\frac{1}{2}, \frac{1}{3}, n\right)$ are the direction cosines of a line, then the value of $n$ is
Answer(a) : $\because\left(\frac{1}{2}, \frac{1}{3}, n\right)$ are the direction cosines of a line
$
\therefore\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^2+n^2=1 \Rightarrow n^2=\frac{23}{36} \Rightarrow n=\frac{ \pm \sqrt{23}}{6}
$
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A line makes the same angle $\theta$ with each of thex and $z$ axis. If the angle $\beta$ which it makes with $y-$axis is such that $\sin^2\beta=3\sin^2\theta$ then $\cos^2\theta$ equals:
AnswerIf a line makes the angle $\alpha,\beta,\gamma$ with $x, y, z$ axix respectively then
$l^2 + m^2 + n^2 = 1$
$\Rightarrow 2l^2 + m^2 = 1$ or $2n^2 + m^2 = 1$
$\Rightarrow2\cos^2\theta=1-\cos^2\beta (\alpha=\gamma=\theta)$
$2\cos^2\theta=\sin^2\beta$
$\Rightarrow2\cos^2\theta=3\sin^2\theta$
$\Rightarrow5\cos^2\theta=3$
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The direction cosines of the line joining (1, -1, 1) and (-1, 1, 1) are:
Answer
- $\frac{1}{\sqrt{2}},- \frac{1}{\sqrt{2}}$
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