2 Marks - Page 2 - Maths STD 12 Science Questions - Vidyadip

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2 Marks

Question 512 Marks

Determine the direction cosines of the normal to the plane and the distance from the origin.
x + y + z

Answer

x + y + z = 1 ...(1)
The direction ratios of normal are 1, 1, and 1
$\therefore\ \sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
Dividing both sides of equation (1) by $\sqrt{3},$ we obtain
$\frac{1}{\sqrt{3}}\text{x}+\frac{1}{\sqrt{3}}\text{y}+\frac{1}{\sqrt{3}}\text{z}=\frac{1}{\sqrt{3}}\ \ .....(2)$
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 are $\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}},\ \text{and}\ \frac{1}{\sqrt{3}}$ and the distance of normal from the origin is $\frac{1}{\sqrt{3}}$ units.

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Question 522 Marks

Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.

Answer

Let P(2, 3, 4), Q(-1, -2, 1), R(5, 8, 7) be given points.
The direction ratios of PQ are -1 - 2, -2 - 3, 1 - 4 i.e. -3, -5, -3
The direction ratios of PR are 5 - 2, 8 - 3, 7 - 4 i.e. 3, 5, 3
Since $\frac{-3}{3}=\frac{-5}{5}=\frac{-3}{3}$
$\therefore$ lines PQ and PR are parallel.
But P is a common point on both the lines points
$\therefore$ P, Q, R are collinear.

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Question 532 Marks

Find the distance of the point (2, 3, 4) from the x-axis.

Answer

A general point (x, y, z) is at a distance $\sqrt{\text{y}^2+\text{z}^2}$ of from the x-axis.
$\therefore$ Distance of the point (2, 3, 4) from x-axis
$=\sqrt{3^2+4^2}=\sqrt{25}$
$=5\text{ units}$.

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Question 542 Marks

Find the vector equation of a plane passing throught a point with position $2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and perpendicular to the vector $4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$

Answer

We know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
Substituting $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{n}}=4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},$ we get
$\vec{\text{r}}\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)=\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{r}}\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)=8-2-3$
$\Rightarrow\vec{\text{r}}\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)=3$

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Question 552 Marks

If a line has direction ratios 2, -1, -2, determine its direction cosines.

Answer

Let the direction cosines of a line be l, m and n.

Now,

$\text{l}=\frac{2}{\sqrt{2^2+(-1)^2+(-2)^2}}=\frac{2}{3}$

$\text{m}=\frac{-1}{\sqrt{2^2+(-1)^2+(-2)^2}}=\frac{-1}{3}$

$\text{n}=\frac{-2}{\sqrt{2^2+(-1)^2+(-2)^2}}=\frac{-2}{3}$

$\therefore$ The direction consines of the line are $\frac{2}{3},\frac{-1}{3},\frac{-2}{3}.$

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Question 562 Marks

Find the equation of the plane passing through the following point:
(0, -1, 0), (3, 3, 0) and (1, 1, 1)

Answer

The equation of the plane passing through points (0, -1, 0), (3, 3, 0) and (1, 1, 1) is given by,
$\begin{vmatrix}\text{x}-0&\text{y}+1&\text{z}-0\\3-0&3+1&0-0\\1-0&1+1&1-0\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-0&\text{y}+1&\text{z}-0\\3&4&0\\1&2&1\end{vmatrix}=0$
$\Rightarrow4\text{x}-3(\text{y}+1)+2\text{z}=0$
$\Rightarrow4\text{x}-3\text{y}+2\text{z}=3$

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Question 572 Marks

Answer each of the following questions in one word or one sentence or as per exact requirement of the quetion:
Find the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane $\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2.$

Answer

The required plane passes through $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ and is parallel to the plane $\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2.$
So, it is normal to the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ which is normal to the given plane.
Hence, the vector equation of the required plane is
$\big[\vec{\text{r}}-\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=0\big]\ \big[\big(\vec{\text{r}}-\vec{\text{a}}\big).\vec{\text{n}}=0\big]$
$\Rightarrow\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{a}+\text{b}+\text{c}$
Thus, the vector equation of the required plane is $\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{a}+\text{b}+\text{c}.$

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Question 582 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 592 Marks

If line makes angle $\alpha,\beta$ and $\gamma$ with the coordinate axes, find the value of $\cos2\alpha+\cos2\beta+\cos2\gamma$.

Answer

It is given that the line makes angles $\alpha,\beta,\gamma$ with the coordinate axis.
$\therefore\text{l}=\cos\alpha,\text{m}=\cos\beta$ and $\text{n}=\cos\gamma$
$\Rightarrow\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\ ......(1)$
Now,
$\cos2\alpha+\cos2\beta+\cos2\gamma$
$=(2\cos^2\alpha-1)+(2\cos^2\beta-1)+(2\cos^2\gamma-1)$
$=2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)-3$
$=2(1)-3$
$=-1$

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Question 602 Marks

Write the ratio in which YZ-plane divides the segment joining P(-2, 5, 9) and Q(3, -2, 4).

Answer

Let the YZ-plane divides the line segment joining points P(-2, 5, 9) and Q(3, -2, 4) in the ratio k : 1.
Using the setion formula, the coordinates of the point of intersection are given by
$\Big(\frac{\text{k}(3)-2}{\text{k}+1},\frac{\text{k}(-2)+5}{\text{k}+1},\frac{\text{k}(4)+9}{\text{k}+1}\Big)$
On the YZ-plane, the X-coordinate of any point is zero.
$\frac{\text{k}(3)-2}{\text{k}+1}=0$
Implies that 3k - 2 = 0
Implies that $\text{k}=\frac{2}{3}$
Thus, the YZ-plane divides the line segment formed by joining the given points in the ratio 2 : 3 internally.

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Question 612 Marks

Find the Cartesian equation of the following plane:

$\vec{\text{r}}.\Big[(\text{s - 2t})\hat{\text{i}}+(3-\text{t})\hat{\text{j}}+(2\text{s + t})\hat{\text{k}}\Big]=15$

Answer

$\vec{\text{r}}.\Big[(\text{s - 2t})\hat{\text{i}}+(3-\text{t})\hat{\text{j}}+(2\text{s + t})\hat{\text{k}}\Big]=15$
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
$\Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\Big).\Big[(\text{s - 2t})\hat{\text{i}}+(3-\text{t})\hat{\text{j}}+(2\text{s + t})\hat{\text{k}}\Big]=15\ \ \ .....(1)$
⇒ (s - 2t)x + (3 - t)y + (2s + t)z = 15
This is the Cartesian equation of the given plane.

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Question 622 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 632 Marks

Write the cartesian and vector equations of y-axis.

Answer

Since y-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 $\vec{\text{b}}=0\hat{\text{i}}+1\hat{\text{j}}+0\hat{\text{k}}$ having direction ratios proportional to 0, 1, 0, the cartesian equation of y-axis is
$\frac{\text{x}-0}{1}=\frac{\text{y}-0}{0}=\frac{\text{z}-0}{0}$
$=\frac{\text{x}}{0}=\frac{\text{y}}{1}=\frac{\text{z}}{0}$
Also, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}+\lambda\big(0\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}\big)$
$=\lambda\hat{\text{j}}$

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Question 642 Marks

Write the value of k for which the planes x - 2y + kz = 4 and 2x + 5y - z = 9 are perpendicular.

Answer

We know that the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are prependicular to each of a₁a₂ + b₁b₂ + c₁c₂ = 0
The given planes are x - 2y + kz = 4 and 2x + 5y - z = 9
⇒ a₁ = 1; b₁ = -2; c₁ = 5; a₂ = 2; b₂ = 5; c₂ = -1
It is given that the given planes are perpendicular.
⇒ a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ (1)(2) + (-2)(5) + (k)(-1) = 0
⇒ 2 - 10 - k = 0
⇒ -8 - k = 0
⇒ k = -8

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Question 652 Marks

In the following cases, find the distance of each of the given points from the corresponding given plane.
Point: (3, -2, 1)
Plane: 2x - y + 2z + 3 = 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 plane is (3, -2, 1) and the plane is 2x - y + 2z + 3 = 0
$\therefore\ \ \text{d}=\Bigg|\frac{2\times3+(-1)\times(-2)+2\times1+3}{\sqrt{(2)^2+(-1)^2+(2)^2}}\Bigg|=\Big|\frac{13}{3}\Big|=\frac{13}{3}$

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Question 662 Marks

Find the equations of the planes that passes through three points.
(1, 1, -1), (6, 4, -5), (-4, -2, 3).

Answer

The given points are A(1, 1, -1), B(6, 4, -5), and C(-4, -2, 3).
$\begin{vmatrix}1&1&-1\\6&4&-5\\-4&-2&3\end{vmatrix}$
= 1(12 - 10) - 1(18 - 20) - 1(-12 + 16)
= 2 + 2 - 4 = 0
Since A, B, C are collinear points, there will be infinite number of
planes passing through the given points.

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Question 672 Marks

Find the vector equation one of following plane.
x + y = 3

Answer

Given, equation of plane is,
x + y = 3
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(\hat{\text{i}}+\hat{\text{j}})=3$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}})=3$
So,
Vector equation of the plane is $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}})=3$

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Question 682 Marks

In the following cases, find the distance of each of the given points from the corresponding given plane.
Point: (2, 3, -5)
Plane: x + 2y - 2z - 9 = 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 plane is (2, 3, -5) and the plane is x + 2y - 2z = 9
$\therefore\ \ \text{d}=\Bigg|\frac{2+2\times3-2(-5)-9}{\sqrt{(1)^2+(2)^2+(-2)^2}}\Bigg|=\frac{9}{3}=3$

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Question 692 Marks

Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).

Answer

The equation of the plane passing through (a, 0, 0), (0, b, 0) and (0, 0, c) is
$\begin{vmatrix}\text{x}-\text{a} & \text{y}-0&\text{z}-0 \\ 0-\text{a} & \text{b}-0 & 0 - 0 \\ 0-\text{a}&0-0&\text{C}-0 \end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-\text{a} & \text{y}&\text{z} \\ -\text{a} & \text{b} & 0 - 0 \\ -\text{a}& 0 &\text{C} \end{vmatrix}=0$
$\Rightarrow\ \text{bc}(\text{x}-\text{a})+\text{acy}+\text{abz}=0$
$\Rightarrow\ \text{bcx}+\text{acy}+\text{abz}=\text{abc}$
Dividing the equationg by abc, we get
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$

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Question 702 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 712 Marks

In the following cases, find the distance of each of the given points from the corresponding given plane.
Point: (0, 0, 0)
Plane: 3x - 4y + 12z = 3

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 (0, 0, 0) and the plane is 3x - 4y + 12z = 3
$\therefore\ \ \text{d}=\Bigg|\frac{3\times0-4\times0+12\times0-3}{\sqrt{(3)^2+(-4)^2+(12)^2}}\Bigg|=\frac{3}{\sqrt{169}}=\frac{3}{13}$

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Question 722 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 732 Marks

Find the value of k so that the lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular to each other.

Answer

Two lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular.
$\frac{\text{x}}{1}=\frac{\text{y}}{-1}=\frac{\text{z}}{\frac{1}{\text{k}}} \ ...(1)$
$\frac{\text{x}-2}{1}=\frac{\text{y}+\frac{1}{2}}{\frac{1}{2}}=\frac{\text{z}-1}{-1} \ ...(2)$
On comparing with $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$
we get,
$\text{x}_1=0, \ \text{y}_1=0, \ \text{z}_1=0$
$\& \ \text{x}_2=2, \ \text{y}_2=\frac{-1}2, \ \text{z}=1$
$\text{a}_1=+1, \ \text{b}_1=-1, \ \text{c}_1=\frac{1}{\text{k}}$
$\& \ \text{a}_2=1, \ \text{b}_2=\frac{1}{2}, \ \text{c}_2=1 $
Since two lines are perpendicular, therefore
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$\Rightarrow1\times1=(-1)\times\frac{1}{2}+\frac{1}{\text{k}}\times(-1)=0$
$\Rightarrow1-\frac{1}{2}-\frac{1}{\text{k}}=0$
$\Rightarrow1-\frac{\text{k}-2}{2\text{k}}=0$
$\Rightarrow2\text{k}-\text{k}-2=0$
$\text{k}=2$

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Question 742 Marks

Write the equation of the plane parallel to the YOZ- plane and passing through (-4, 1, 0).

Answer

The equation of the plane parallel ot the plane YOZ is x = b .....(i), where b is a constant. It is given that plane passes throught (-4, 1, 0). So, -4 = b
Substituting this value in (i), we get x = -4, which is the required equation of the plane.

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Question 752 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 762 Marks

Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y − z = 7.

Answer

Let the equation of a plane parallel to the given plane be
3x + 2y - z = k ....(1)
This passes through (2, -1, 1).
So, 3(2) + 2(-1) - (1) = k
k = 3
Substituting this in(1),
We get,
3x + 2y - z = 3, which is the equation of the required plane.

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Question 772 Marks

Find the Cartesian equation of the following plane:

$\vec{\text{r}}.\Big(2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}\Big)=1$

Answer

$\vec{\text{r}}.\Big(2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}\Big)=1\ \ \ ....(1)$
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
$\Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}-\text{z}\hat{\text{k}}\Big).\Big(2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}\Big)=1$
⇒ 2x + 3y - 4z = 1
This is the Cartesian equation of the plane.

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Question 782 Marks

If a line makes angles of 90°, 60° and 30° with the positive direction of x, y, and z-axis respectively, find its direction cosines.

Answer

Let l, m and n be the direction cosines of a line.

$\text{l}=\cos90^{\circ}=0$

$\text{m}=\cos60^{\circ}=\frac{1}{2}$

$\text{n}=\cos30^{\circ}=\frac{\sqrt{3}}{2}$

$\therefore$ The direction consines of the line are $0,\frac{1}{2},\frac{\sqrt{3}}{2}.$

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Question 792 Marks

If a line makes angles 90° and 60° respectively with the positive direction of x and y axes, find the angle which it makes with the positive direction of z-axis.

Answer

Let the direction cosines of the line be l, m and n.
We know that l² + m² + n² = 1
Let the line make angle $\theta$ with positive direction of the z-axis
$\alpha=90^\circ,\beta=60^\circ,\gamma=\theta$.
So, $\cos^290^\circ+\cos^260^\circ+\cos^2\theta=0$
$\Rightarrow0+\Big(\frac{1}{2}\Big)^2+\cos^2\theta=1$
$\Rightarrow\cos^2\theta=1-\frac{1}{4}$
$\Rightarrow\cos^2\theta=\frac{3}{4}$
$\Rightarrow\cos\theta=\pm\frac{\sqrt{3}}{4}$
$\Rightarrow\theta=30^\circ$or $150^\circ$

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Question 802 Marks

Determine the direction cosines of the normal to the plane and the distance from the origin.
z = 2

Answer

The equation of the plane is z = 2 or 0x + 0y + z = 2 ...(1)
The direction ratios of normal are 0, 0, and 1.
$\therefore\ \sqrt{0^2+0^2+1^2}=1$
Dividing both sides of equation (1) by 1, we obtain
0.x + 0.y + 1.z = 2
This 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 the perpendicular drawn from the origin.
Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

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Question 812 Marks

Answer the following quations in one word or one sentence or as per exact requirement of the question:
Write the distance of a point P(a, b, c) from x-axis.

Answer

We know that a general point (x, y, z) has distance $\sqrt{\text{y}^2+\text{z}^2}$
$\therefore$ Distance of a point P(x, y, z) from x-axis $=\sqrt{\text{b}^2+\text{c}^2}$.

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Question 822 Marks

Write the equation of the plane corntaining the lines $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ and $\vec{\text{r}}=\vec{\text{a}}+\mu\vec{\text{c}}$.

Answer

The given equation of the plane is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ and $\vec{\text{r}}=\vec{\text{a}}+\mu\vec{\text{c}}$.
So, the plane passes through the vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ and $\vec{\text{c}}$.
So, the plane passes through the vector $\vec{\text{a}}$ whose normal vector is $\vec{\text{b}}\times\vec{\text{a}}$
(It means that $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{a}}$)
So, the eqution of the plane in scalar product from is
$(\vec{\text{r}}-\vec{\text{a}}).\vec{\text{n}}=0$
$\Rightarrow(\vec{\text{r}}-\vec{\text{a}}).(\vec{\text{b}}\times\vec{\text{c}})=0$

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Question 832 Marks

Find the cartesian form of the equation of a plane whose vector equation is:
$\vec{\text{r}}\cdot\big(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)=9$

Answer

Here, equation of the plane is,
$\vec{\text{r}}\cdot\big(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)=9$
Let $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}},$ then
$\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)\big(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)=9$
$(\text{x})(-1)+(\text{y})(1)+(\text{z})(2)=9$
$-\text{x}+\text{y}+2\text{z}=9$
Cartesian form of the equation of the plane is,
$-\text{x}+\text{y}+2\text{z}=9$

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Question 842 Marks

Find the value of k so that the lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular to each other.

Answer

Two lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular.
$\frac{\text{x}}{1}=\frac{\text{y}}{-1}=\frac{\text{z}}{\frac{1}{\text{k}}} \ ...(1)$
$\frac{\text{x}-2}{1}=\frac{\text{y}+\frac{1}{2}}{\frac{1}{2}}=\frac{\text{z}-1}{-1} \ ...(2)$
On comparing with $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$
we get,
$\text{x}_1=0, \ \text{y}_1=0, \ \text{z}_1=0$
$\& \ \text{x}_2=2, \ \text{y}_2=\frac{-1}2, \ \text{z}=1$
$\text{a}_1=+1, \ \text{b}_1=-1, \ \text{c}_1=\frac{1}{\text{k}}$
$\& \ \text{a}_2=1, \ \text{b}_2=\frac{1}{2}, \ \text{c}_2=1 $
Since two lines are perpendicular, therefore
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$\Rightarrow1\times1=(-1)\times\frac{1}{2}+\frac{1}{\text{k}}\times(-1)=0$
$\Rightarrow1-\frac{1}{2}-\frac{1}{\text{k}}=0$
$\Rightarrow1-\frac{\text{k}-2}{2\text{k}}=0$
$\Rightarrow2\text{k}-\text{k}-2=0$
$\text{k}=2$

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Question 852 Marks

Find the vector equation of a plane which is at a distance of 5 unit from the origin and which is normal to the vector $\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$

Answer

It is given that the normal vector, $\vec{\text{n}}=\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
Now, $\text{n}=\frac{\vec{\text{n}}}{|\vec{\text{n}}|}$
$=\frac{\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}}{\sqrt{1+4+4}}$
$=\frac{\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}}{3}$
$=\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}$
The equation of a plane in normal form is
$\vec{\text{r}}\cdot{\text{n}}={\text{d}}$ (where d is the distance of the plane from the origin)
Substituting $\text{n}=\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}$ and d = 5
Here,
$\vec{\text{r}}\cdot\Big(\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}\Big)=5$

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Question 862 Marks

Write the cartesian and vector equations of z-axis.

Answer

Since z-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 $\vec{\text{b}}=0\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}$ having direction ratios proportional to 0, 0, 1, the cartesian equation of z-axis is
$\frac{\text{x}-0}{0}=\frac{\text{y}-0}{0}=\frac{\text{z}-0}{1}$
$=\frac{\text{x}}{0}=\frac{\text{y}}{0}=\frac{\text{z}}{1}$
Also, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}+\lambda\big(0\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}\big)$
$=\lambda\hat{\text{k}}$

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Question 872 Marks

Find the cartesian form of the equation of a plane whose vector equation is:
$\vec{\text{r}}\cdot\big(12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+5=0$

Answer

Given the vector equation of a plane,
$\vec{\text{r}}\cdot\big(12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+5=0$
Let, $\vec{\text{r}}=\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)$
$\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)\big(12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+5=0$
$(\text{x})(12)+(\text{y})(-3)+(\text{z})(4)=0$
$12\text{x}-3\text{y}+4\text{z}+5=0$
Cartesian form of the equation of the plane of the plane is given by
$12\text{x}-3\text{y}+4\text{z}+5=0$

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Question 882 Marks

Find the intercepts cut off by the plane 2x + y - z = 5.

Answer

2x + y - z = 5 ....(1)
Dividing both sides of equation (1) by 5, we obtain
$\frac{2}{5}\text{x}+\frac{\text{y}}{5}-\frac{\text{z}}{5}=1$
$\Rightarrow\frac{\text{x}}{\frac{5}{2}}\text{x}+\frac{\text{y}}{5}+\frac{\text{z}}{-5}=1\ \ ...(2)$
It is known that the equation of a plane in intercept form $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1,$
where a, b, c are the intercepts cut off by the plane at x, y, and z axes respectively.
Therefore, for the given equation,
$\text{a}=\frac{5}{2},\ \text{b}=5,\ \text{and}\ \text{c}=-5$
Thus, the intercepts cut off by the plane are $\frac{5}{2},\ 5,\ \text{and}\ -5.$

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Question 892 Marks

Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.

Answer

For x-intercepts, put y = 0 and z = 0 in the given eqution.
Then, we get
2x + 0 - 0 = 5
⇒ 2x = 5
$\Rightarrow\text{x}=\frac{5}{2}$

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Question 902 Marks

Find the value of k so that the lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular to each other.

Answer

Two lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular.
$\frac{\text{x}}{1}=\frac{\text{y}}{-1}=\frac{\text{z}}{\frac{1}{\text{k}}} \ ...(1)$
$\frac{\text{x}-2}{1}=\frac{\text{y}+\frac{1}{2}}{\frac{1}{2}}=\frac{\text{z}-1}{-1} \ ...(2)$
On comparing with $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$
we get,
$\text{x}_1=0, \ \text{y}_1=0, \ \text{z}_1=0$
$\& \ \text{x}_2=2, \ \text{y}_2=\frac{-1}2, \ \text{z}=1$
$\text{a}_1=+1, \ \text{b}_1=-1, \ \text{c}_1=\frac{1}{\text{k}}$
$\& \ \text{a}_2=1, \ \text{b}_2=\frac{1}{2}, \ \text{c}_2=1 $
Since two lines are perpendicular, therefore
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$\Rightarrow1\times1=(-1)\times\frac{1}{2}+\frac{1}{\text{k}}\times(-1)=0$
$\Rightarrow1-\frac{1}{2}-\frac{1}{\text{k}}=0$
$\Rightarrow1-\frac{\text{k}-2}{2\text{k}}=0$
$\Rightarrow2\text{k}-\text{k}-2=0$
$\text{k}=2$

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Question 912 Marks

Write the equation of the plane $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ in scalar product from.

Answer

The given equation of the plane is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$
So, the plane passes through the vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ and $\vec{\text{c}}$.
So, the plane passes through the vector $\vec{\text{a}}$ whose normal vector is $\vec{\text{b}}\times\vec{\text{a}}$
(It means that $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{a}}$)
So, the eqution of the plane in scalar product from is
$(\vec{\text{r}}-\vec{\text{a}}).\vec{\text{n}}=0$
$\Rightarrow(\vec{\text{r}}-\vec{\text{a}}).(\vec{\text{b}}\times\vec{\text{c}})=0$

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Question 922 Marks

Write the coordinates of the projection of point P(2, -3, 5) on Y-axis.

Answer

The coordinates of the projection of the point P(2, -3, 5) on Y-axis are (0, -3, 0) as both x and z coordinates of each point on the y-axis are equal to zero.

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Question 932 Marks

Write the coordinates of the projection of point P(x, y, z) on XOZ-plane.

Answer

The projection of the point P(x, y, z) on XOZ-plane is (x, 0, z) as Y-coordinates of any point on XOZ-plane are equal to zero.

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Question 942 Marks

What are the direction cosines of Y-axis?

Answer

The y-axis makes angles 90°, 0° 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. 0, 1, 0.

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