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Three Dimensional Geometry · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 94 questions.

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Question 12 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.
Answer
The 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.
$\Rightarrow\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}=0$
$\Rightarrow\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)$
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Question 22 Marks
Write the direction consines of the line whose cartesian equations are 2x = 3y = -z.
Answer
We have

2x = 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

$\frac{3}{\sqrt{3^2+2^2+(-6)^2}},\frac{2}{\sqrt{3^2+2^2+(-6)^2}},\frac{-6}{\sqrt{3^2+2^2+(-6)^2}}$

$=\frac{3}{7},\frac{2}{7},-\frac{6}{7}$

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Question 32 Marks
Write the distance of the point P(x, y, z) from XOY plane.
Answer
The distance of point P(x, y, z) from the XOY plane is z.
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Question 42 Marks
Find the vector equation of the plane which is at a distance of 5 units from the orgin and its normal vector is $2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}.$
Answer
Given:

Normal vector, $\hat{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$

Perpendicular distance, d = 5 units

The vector equation of a plane that is at a distance of 5 units from the origin and has its normal vector $\hat{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$ is as follows:

$\vec{\text{r}}.\hat{\text{n}}=\text{d}$

$\vec{\text{r}}.\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)=5.$

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Question 52 Marks
Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, -1), (4, 3, -1).
Answer
We know that direction ratios of the line joining the origin (0, 0, 0) to the point are

x2 - x1, y2 - y1, z2 - z1 = 2 - 0, 1 - 0, 1 - 0 = 2, 1, 1 = a1, b1, c1

Similarly, direction ratios of the line joining the points (3, 5, -1) and (4, 3, -1) are

x2 - x1, y2 - y1, z2 - z1 = 4 - 3, 3 - 5, -1 - (-1) = 1, -2, 0 = a2, b2, c2

For these two lines, a1a2 + b1b2 + c1c2

= 2(1) + 1(-2) + 1(0) = 2 - 2 + 0 = 0

Therefore, the two given lines are perpendicular to each other.

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Question 62 Marks
If the equations of a line AB are $\frac{3-\text{x}}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-5}{4},$ write the direction ratios of a line parallel to AB.
Answer
We have

$\frac{3-\text{x}}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-5}{4}$

The equation of the line AB can be re-written as

$\frac{\text{x}-3}{-1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-5}{4}$

Thus, the direction ratios of the line parllel to Ab are proportional to -1, -2, 4.

 

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Question 72 Marks
Find the equation of a line parallel to x-axis and passing through the origin.
Answer
Vector equation of a line is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
The direction cosines of the x-axis are (1, 0, 0). Equation of a line parallel to the x-axis and passing through the origin is
$\vec{\text{r}}=\big(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)+\lambda\big(1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)$
$\vec{\text{r}}=\lambda\hat{\text{i}}$
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Question 82 Marks
The equations of a line are given by $\frac{4-\text{x}}{3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}.$ Write the direction cosines of a line parallel to this line.
Answer
We have

$\frac{4-\text{x}}{3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}$

The equation of the given line can be re-written as.

$\frac{\text{x}-4}{-3}=\frac{\text{y}+3}{3}=\frac{\text{z}+2}{6}$

The direction ratios of the line parallel to the given line are proportional to -3, 3, 6.

Hence, the direction cosines of the line parallel to the given line are proportional to

$\frac{-3}{\sqrt{(-3)^2+3^2+6^2}},\frac{-3}{\sqrt{(3)^2+3^2+6^2}},\frac{6}{\sqrt{(-3)^2+3^2+6^2}}$

$=\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}$

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Question 92 Marks
Write the vector equation of a line passing through a point having position vector $\vec{\alpha}$ and parallel to vector $\vec{\beta}.$
Answer
The vector equation of the line passing the point having position vector $\vec{\alpha}$ and parallel to vector $\vec{\beta}$is $\vec{\text{r}}=\vec{\alpha}+\lambda\vec{\beta}.$
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Question 102 Marks
Write the formula for the shortest distance between the lines

$\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}.$

Answer
The shortest distance d between the parallel lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}$ is given by
$\text{d}=\frac{\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|}{\big|\vec{\text{b}}\big|}$
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Question 112 Marks
Write the distances of the point (7, -2, 3) from XY, YZ and XZ-planes.
Answer
The 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.
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Question 122 Marks
Write the vector equation of a line given by $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}.$
Answer
We have
$\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}$
The given line passes through the point (5, -4, 6) and has direction ratios proportional to 3, 7, 2.
Vector equation of the given line passing through the point having position vector $\vec{\text{a}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}$ and parallel to a vector $\vec{\text{b}}=3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}$ is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\vec{\text{r}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}+\lambda\big(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}\big)$
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Question 132 Marks
Find the equation of the line passing through the points (2, -1, 3) and parallel to the line $\vec{\text{r}}=\big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}\big).$
Answer
The 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 $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}\big)$
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Question 142 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.
Answer
Here, 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,
$\vec{\text{r}}\cdot\hat{\text{k}}=3$
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Question 152 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 $\vec{\text{r}}=7\hat{\text{i}}-6\hat{\text{k}}+\mu\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big).$
Answer
Let $\theta$ 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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Question 162 Marks
Find the Cartesian equations of the line which passes through the point (-2, 4, -5) and is parallel to the line $\frac{\text{x}+3}{3}=\frac{4-\text{y}}{5}=\frac{\text{z}+8}{6}.$
Answer
The 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 $\frac{\text{x}+2}{3}=\frac{\text{y}-4}{-5}=\frac{\text{z}+5}{6}.$

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Question 172 Marks
Write the equation of the plane parallel to XOY- plane and passing through the point (2, -3, 5).
Answer
The equation of the plane parallel to the plane XOY is z = b .....(i), where b is a constant. It is given that this palne passes through (2, -3, 5).
So, 5 = b
Substituting this value in (i), we get z = 5, which is the required equation of the plane.
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Question 182 Marks
Find the vector equation one of following plane.
x + y - z = 5
Answer
Given, equation of plane is,
x + y - z = 5
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})=5$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=5$
So,
Vector equation of the plane is $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=5$
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Question 192 Marks
Find the vector equation one of following plane.
2x - y + 2z = 8
Answer
Given, equation of plane is,
2x - y + 2z = 8
 $(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=8$
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=8$
So,
Vector equation of the plane is $\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=8$
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Question 202 Marks
Find the equation of the plane passing through the following point:
(2, 3, 4), (-3, 5, 1) and (4, -1, 2)
Answer
The equation of the plane passing through points (2, 3, 4), (-3, 5, 1) and (4, -1, 2) is given by,
$\begin{vmatrix}\text{x}-2&\text{y}-3&\text{z}-4\\-3-2&5-3&1-4\\4-2&-1-3&2-4\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-3&\text{z}-4\\-5&2&-3\\2&-4&-2\end{vmatrix}=0$
$\Rightarrow-16(\text{x}-2)-16(\text{y}-3)+16(\text{z}-4)=0$
$\Rightarrow(\text{x}-2)+(\text{y}-3)-(\text{z}-4)=0$
$\Rightarrow\text{x}+\text{y}-\text{z}=1$
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Question 212 Marks
Write the cartesian and vector equations of x-axis.
Answer
Since x-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 $\vec{\text{b}}=1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ having direction ratios proportional to 1, 0, 0, the cartesian equation of x-axis is
$\frac{\text{x}-0}{1}=\frac{\text{y}-0}{0}=\frac{\text{z}-0}{0}$
$=\frac{\text{x}}{1}=\frac{\text{y}}{0}=\frac{\text{z}}{0}$
Also, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}+\lambda\big(\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)$
$=\lambda\hat{\text{i}}$
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Question 222 Marks
What are the direction cosines of Z-axis?
Answer
The z-axis makes angles 90°, 90° and 0° with x, y and z axes, respectively.
Therefore, the direction cosines of x-axis are cos 90°, cos 90°, cos 0°, i.e. 0, 0, 1.
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Question 232 Marks
Write the value of k for which the line $\frac{\text{x}-1}{2}=\frac{\text{y}-1}{3}=\frac{\text{z}-1}{\text{k}}$ is perpendicular to the normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})=4.$
Answer
Direction ratios of the given line $\frac{\text{x}-1}{2}=\frac{\text{y}-1}{3}=\frac{\text{z}-1}{\text{k}}$ are proportional to 2, 3, k.
Direction ratios of the normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})=4$ are 2, 3, 4.
Given that these two are perpendicular.
⇒ (2) (2) + (3) (3) + (k) (4) (Because a1a2 + b1b2 + c1c2 = 0)
⇒ 4 + 9 + 4k = 0
⇒ 13 + 4k = 0
$\Rightarrow\text{k}=\frac{-13}{4}$.
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Question 242 Marks
If a line has direction ratios proportional to 2, -1, -2, then what are its direction consines?
Answer
If a line has direction ratios proportional to 2, -1 and -2, then its direction cosines are
$\frac{2}{\sqrt{(2)^2+(-1)+(-2)^2}},\frac{-1}{\sqrt{(2)^2+(-1)+(-2)^2}},\frac{-2}{\sqrt{(2)^2+(-1)+(-2)^2}}$
$=\frac{2}{3},-\frac{1}{3},-\frac{2}{3}$
Thus, the direction cosines are $=\frac{2}{3},-\frac{1}{3},-\frac{2}{3}$.
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Question 252 Marks
Find the equation of the plane passing through the following point:
(1, 1, 1), (1, -1, 2) and (-2, -2, 2)
Answer
The equation of the plane passing through points (1, 1, 1), (1, -1, 2) and (-2, -2, 2) is given by,
$\begin{vmatrix}\text{x}-1&\text{y}-1&\text{z}-1\\1-1&-1-1&2-1\\-2-1&-2-1&2-1\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-1&\text{y}-1&\text{z}-1\\0&-2&1\\-3&-3&1\end{vmatrix}=0$
$\Rightarrow1(\text{x}-1)-3(\text{y}-1)-16(\text{z}-1)=0$
$\Rightarrow\text{x}-1-3\text{y}+3-6\text{z}+6=0$
$\Rightarrow\text{x}-3\text{y}-6\text{z}+8=0$
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Question 262 Marks
If a line makes angles 90º, 135º, 45º with the x, y and z-axes respectively, find its direction cosines.
Answer
Let l, m, n be the direction cosines of the line with direction angles 90º, 135º, 45º.
$\therefore\text{l}=\cos90^\circ,\ \text{m}=\cos135^\circ=\cos(180^\circ-45^\circ)=-\cos45^\circ=-\frac{1}{\sqrt{2}},$
$\text{n}=\cos45^\circ=\frac{1}{\sqrt{2}}$
$\therefore$ direction cosines are $0,-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}.$
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Question 272 Marks
Determine the direction cosines of the normal to the plane and the distance from the origin.
2x + 3y - z = 5
Answer
2x + 3y - z = 5 ...(1)
The direction ratios of normal are 2, 3, and -1.
$\therefore\ \sqrt{(2)^2+(3)^2+(-1)^2}=\sqrt{14}$
Dividing both sides of equation (1) by $\sqrt{14},$ we obtain
$\frac{2}{\sqrt{14}}\text{x}+\frac{3}{\sqrt{14}}\text{y}-\frac{1}{\sqrt{14}}\text{z}=\frac{5}{\sqrt{14}}$
This equation 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 normal from the origin.
Therefore, the direction cosines of the normal to the plane are $\frac{2}{\sqrt{14}},\ \frac{3}{\sqrt{14}},\ \text{and}\ \frac{-1}{\sqrt{14}}$ and the distance of normal from the origin is $\frac{5}{\sqrt{14}}$ units.
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Question 282 Marks
Find the length of the perpendicular drawn from the origin to the plane 2x − 3y + 6z + 21 = 0.
Answer
We know that the distance of the point (x1, y1, z1) from the plane ax + by + cz + d = 0 is given by
$\frac{|\text{ac}_1+\text{by}_1+\text{cz}_1+\text{d}|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|2(0)-3(0)+6(0)+21|}{\sqrt{2^2+(-3)^2+6^2}}$
$=\frac{|21|}{\sqrt{4+9+36}}$
$=\frac{21}{7}$
$=3\text{ units}$
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Question 292 Marks
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}.$
Answer
The normal vector is, $\vec{\text{n}}=3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}$
$\therefore\ \ \hat{\text{n}}=\frac{\vec{\text{n}}}{|\vec{\text{n}}|}=\frac{3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}}{\sqrt{(3)^2+(5)^2+(6)^2}}=\frac{3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}}{\sqrt{70}}$
It is known that the equation of the plane with position vector $\vec{\text{r}}$ is given by,
$\vec{\text{r}}.\hat{\text{n}}=\text{d}$
$\Rightarrow\ \vec{\text{r}}.\Bigg(\frac{3\hat{\text{i}}+5\hat{\text{j}}-6\hat{\text{k}}}{\sqrt{70}}\Bigg)=7$
This is the vector equation of the required plane.
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Question 302 Marks
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line $\text{5x – 25 = 14 – 7y = 35z.}$
Answer
Equation 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
$\vec{\text{r}} = (\hat{\text{i}} + 2\hat{\text{j}} - \hat{\text{k}}) + \lambda (7\hat{\text{i}} - 5\hat{\text{j}} + \hat{\text{k}})$
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Question 312 Marks
Write the angle between the line $\frac{\text{x}-1}{2}=\frac{\text{y}-2}{1}=\frac{\text{z}+3}{-\text{2}}$ and the plane x + y + 4 = 0.
Answer
The given line is parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$ and the given plane is normal to the vector $\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$.
We know that the angle $\theta$ between the line and the plane is given by
$\sin\theta=\frac{\vec{\text{b}}.\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
$=\frac{\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)\big(\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}\big)}{|\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}||\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}|}$
$=\frac{1+2+0}{\sqrt{1+4+4}\sqrt{1+1+0}}$
$=\frac{3}{3\sqrt{2}}$
$=\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\sin^{-1}\Big(\frac{1}{\sqrt{2}}\Big)=45^\circ$
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Question 322 Marks
Determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0
Answer
5y + 8 = 0
0x - 5y + 0z = 8 ....(1)
The direction ratios of normal are 0, -5, and 0.
$\therefore\ \sqrt{0+(-5)+0}=5$
Dividing both sides of equation (1) by 5, we obtain
$-\text{y}=\frac{8}{5}$
This equation 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 normal from the origin.
Therefore, the direction cosines of the normal to the plane are 0, -1, and 0 and the distance of normal from the origin is $\frac{8}{5}$ units.
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Question 332 Marks
Write the distance of the point (3, −5, 12) from X-axis?
Answer
The distance of a general point (x, y, z) from x-axis is $\sqrt{\text{y}^2+\text{z}^2}$.
$\therefore$ Distance of the point (3, -5, 12) from x-axis $=\sqrt{(-5)^2+12^2}$
$=\sqrt{169}$
= $13\text{ units}$.
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Question 342 Marks
What are the direction cosines of X-axis?
Answer
The x-axis makes angles 0°, 90° 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. 1, 0, 0.
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Question 352 Marks
The x-coordinate of a point on the line joining the points P(2, 2, 1) and Q(5, 1, – 2) is 4. Find its z-coordinate.
Answer
Equation of line PQ is $\frac{\text{x - 2}}{3} = \frac{\text{y - 2}}{-1} = \frac{\text{z - 1}}{-3}$

Any point on the line is $(3\lambda + 2, -\lambda + 2, -3\lambda + 1)$

$3\lambda + 2 = 4 \Rightarrow \lambda = \frac{2}{3} \therefore \text{z coord.} = -3\bigg(\frac{2}{3}\bigg) + 1 = -1.$

Alternate Answer

Let R(4, y, z) lying on PQ divides PQ in the ratio k:1

$\Rightarrow 4 = \frac{\text{5k + 2}}{\text{k + 1}} \Rightarrow \text{k = 2}.$

$\therefore \text{z} = \frac{2(-2) + 1 (1)}{2 + 1} = \frac{-3}{3} = -1.$

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Question 362 Marks
Write direction cosines of a line parallel to z-axis.
Answer
A line parallel to z-axis, makes an angle of 90°, 90° and 0° with the x, y, and z axes, respectively.
Thus, the direction cosines are given by
$\text{l}=\cos90^\circ=0$
$\text{m}=\cos90^\circ=0$
$\text{n}=\cos0^\circ=1$
Therefore, direction cosines of a line parallel to the z-axis 0, 0, 1.
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Question 372 Marks
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point: (-6, 0, 0)
Plane: 2x - 3y + 6z - 2 = 0
Answer
It is known that the distance between a point p(x1, y1, z1), 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 (-6, 0, 0) and the plane is 2x - 3y + 6z - 2 = 0
$\therefore\ \ \Bigg|\frac{2(-6)-3\times0+6\times0-2}{\sqrt{(2)^2+(-3)^2}+(6)^2}\Bigg|=\Big|\frac{-14}{\sqrt{49}}\Big|=\frac{14}{7}=2.$
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Question 382 Marks
Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
Answer
The equation of the plane ZOX is
y = 0
Any plane parallel to it is of the form, y = a
Since the y-intercept of the plane is 3,
$\therefore$ a = 3
Thus, the equation of the required plane is y = 3.
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Question 392 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.
Answer
Vector 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: $\frac{\text{x - 2}}{3} = \frac{\text{y + 3}}{4} = \frac{\text{z - 4}}{-5}$
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Question 402 Marks
Define direction cosines of a direction line.
Answer
The direction cosines of a direction line segment are the cosines of the direction angles of the line segment. Let two points A(x1, y1, z1) and (x2, y2, z2) define the directed line segment AB.
The direction cosines of AB are given by cos
$\alpha=\frac{\text{x}_2-\text{x}_1}{\text{d}}$
$\cos\beta=\frac{\text{y}_2-\text{y}_1}{\text{d}}$
$\cos\gamma=\frac{\text{z}_2-\text{z}_1}{\text{d}}$
Here, d is the distance between A and B.
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Question 412 Marks
Show that the line through the points (1, -1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Answer
We know that direction ratios of the line joining the points A(1, -1, 2) and B(3, 4, -2) are
x2 - x1, y2 - y1, z2 - z1
⇒ 3 - 1, 4 - (-1), -2 - 2
⇒ 2, 5, -4 = a1, b1, c1
Again, direction ratios of the line joining the points C(0, 3, 2) and D(3, 5, 6) are
x2 - x1, y2 - y1, z2 - z1
⇒ 3 - 0, 5 - 3, 6 - 2
⇒ 3, 2, 4 = a2, b2, c2 (say)
For lines AB and CD, a1a2 + b1b2 + c1c2
= 2 × 3 + 5 × 2 + (-4) × 4 = 6 + 10 - 16 = 0
Therefore, line AB is perpendicular to line CD.
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Question 422 Marks
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line $\text{5x – 25 = 14 – 7y = 35z.}$
Answer
Equation 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
$\vec{\text{r}} = (\hat{\text{i}} + 2\hat{\text{j}} - \hat{\text{k}}) + \lambda (7\hat{\text{i}} - 5\hat{\text{j}} + \hat{\text{k}})$
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Question 432 Marks
Write the distance of the plane $\vec{\text{r}}.(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=12$ form the origin.
Answer
The given equation of the plane is
$\vec{\text{r}}.(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=12$ or $\vec{\text{r}}.\vec{\text{n}}=-6$, where $\vec{\text{n}}=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{4+1+4}=3$
For reducing the given equation to normal form, we need to divide both sides by $|\vec{\text{n}}|$.
Then, we get $\vec{\text{r}}.\frac{\vec{\text{n}}}{|\vec{\text{n}}|}=\frac{12}{|\vec{\text{n}}|}$
$=\vec{\text{r}}.\Big(\frac{2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}}{3}\Big)=\frac{12}{3}$
$\Rightarrow\vec{\text{r}}.\Big(\frac{2}{3}\hat{\text{i}}+\frac{1}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}\Big)=4,\ ...(1)$
The eqution of the plane normal from is.
$\vec{\text{r}}.\text{n}=\text{d}\ ....(2)$
(where d is the distance of the plane from the origin)
Comparing (1) and (2),
Length of the perpendicular from the origin to the plane = d = 4 units.
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Question 442 Marks
Write a vector normal to the plane $\vec{\text{r}}=\text{l}\vec{\text{b}}+\text{m}\vec{\text{c}}$ .
Answer
The equation of the given plane is
$\vec{\text{r}}=\text{l}\vec{\text{b}}+\text{m}\vec{\text{c}}$
So, the plane passes parallel to the vectors $\vec{\text{b}}$ and $\vec{\text{c}}$.
So, the vector normal to the plane is $\vec{\text{b}}\times\vec{\text{c}}$.
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Question 452 Marks
Write the coordinate axis to which the line  $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-1}{0}$ is perpendicular.
Answer
We have
$\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-1}{0}$
The given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}+0\hat{\text{k}}$
Let $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ be perpendicular 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.
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Question 462 Marks
Write the position vector of the point where the line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$  meets the plane $\vec{\text{r}}.\vec{\text{n}}=0$.
Answer
The given equation of the plane is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\ ....(1)$
Given equation of the plane is
$\vec{\text{r}}.\vec{\text{n}}=0$
$\Rightarrow\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big)\vec{\text{n}}=0$ [ From (1) ]
$\Rightarrow\vec{\text{a}}.\vec{\text{n}}+\lambda\vec{\text{b}}.\vec{\text{n}}=0$
$\Rightarrow\lambda=-\Big(\frac{\vec{\text{a}}.\vec{\text{n}}}{\vec{\text{b}}.\vec{\text{n}}}\Big)$
Substituting this in (1), we get
$\vec{\text{r}}=\vec{\text{a}}-\Big(\frac{\vec{\text{a}}.\vec{\text{n}}}{\vec{\text{b}}.\vec{\text{n}}}\Big)\vec{\text{b}},$ which is the required position vector that lies both on the line and the plane.
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Question 472 Marks
Write the general equation of a plane parallel to X-axis.
Answer
The general equation of a plane is
ax + by + cz + d = 0 .....(i)
This plane is parallel to the X-axis.
It means that this plane passes through the point (0, y, z). So,
a(0) + by + cz + d = 0
⇒ by + cz + d = 0
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Question 482 Marks
Find the Cartesian equation of the following plane:
$\vec{\text{r}}.\Big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\Big)=2$​​​​​
Answer
It is given that equation of the plane is
$\vec{\text{r}}.\Big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\Big)=2\ \ \ ....(1)$
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
$\Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\Big).\Big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\Big)=2$
⇒ x + y - z = 2
This is the Cartesian equation of the plane.
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Question 492 Marks
Write the ratio in which the line segment joining (a, b, c) and (-a, -b, -c) is divided by the xy-plane.
Answer
Suppose 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.
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Question 502 Marks
Write the inclination a line with z-axis, if its direction ratios are proportional to 0, 1, -1.
Answer
We know that if a line has direction ratio (a, b, c), then the cosine of its angle with the z-axis is given by
$\cos\gamma=\frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
Suppose the inclination of the line with direction ratio (0, 1, -1) with z-axis is $\gamma$.
Now,
$\cos\gamma=\frac{1}{\sqrt{0+1+1}}$
$=-\frac{1}{\sqrt{2}}$
Implies that $\lambda=\frac{3\pi}{4}$
Hence, the inclination of the line with z-axis is $\frac{3\pi}{4}$.
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2 Marks - Maths STD 12 Science Questions - Vidyadip