3 Marks

Question 513 Marks

Write the ratio in which the plane 4x + 5y − 3z = 8 divides the line segment joining the points (−2, 1, 5) and (3, 3, 2).

Answer

We know that the ratio in which the plane ax + by + cz + d = 0 divides the line sebment joining
(x₁, y₁, z₁) and (x₂, y₂, z₂) is $\frac{-(\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d})}{\text{ax}_2+\text{by}_2+\text{cz}_2+\text{d}}$
Here, a = 4,b = 5,c = -3,d = -8,x₁ = -2,y₁ = 1,z₁ = 5,x₂ = 3,y₂ = 3,z₂ = 2
So, the required ratio
$=\frac{-(4(-2)+5(1)-3(5)-8)}{4(3)+5(3)-3(2)-8}$
$=\frac{-(-8+5-15-8)}{12+15-6-8}$
$=\frac{26}{13}$
$=\frac{2}{1}$ or $2 :1.$

View full question & answer→

Question 523 Marks

Reduce the equation 2x - 3y - 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.

Answer

Given equation of plane is,
2x - 3y - 6z = 14
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}})=14$
Dividing the equation by $\sqrt{(2)^2+(-3)^2+(-6)^2}$
$\vec{\text{r}}\cdot\frac{(2\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}})}{\sqrt{4+9+36}}=\frac{14}{\sqrt{4+9+36}}$
$\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}-\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=\frac{14}{7}$
$\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}-\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=2\ ...(\text{i})$
We know that the vector equation of a plane with distance d from origin and normal to unit vector $\hat{\text{n}}$ is given by
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{d}}\ ...(\text{ii})$
Comparing (i) and (ii),
d = 2 and
$\hat{\text{n}}=\frac{2}{7}\hat{\text{i}}-\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}$
So, distance of plane from origin = 2 unit
Direction cosine of normal to plane $=\frac{2}{7},-\frac{3}{7},\frac{6}{7}$

View full question & answer→

Question 533 Marks

Write the distance between the parallel planes 2x − y + 3z = 4 and 2x − y + 3z = 18.

Answer

The given equation are
2x − y + 3z = 4 ....(1)
The second equation of the plane is
2x − y + 3z = 18 .....(2)
We know that distance between two planes ax + by + cz = d₁ and ax + by + cz = d₂ is $\frac{|\text{d}_2-\text{d}_1|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance is
$\frac{|18-4|}{\sqrt{2^2+(-1)^2+3^2}}$
$=\frac{|14|}{\sqrt{4+1+9}}$
$=\frac{14}{\sqrt{14}}$
$=\sqrt{14}\text{ units}$

View full question & answer→

Question 543 Marks

Find the equation of the plane with intercept 3 on the y-axis and parallel to the ZOX plane.

Answer

The equation of the plane parallel to the plane ZOX is,
y = b ....(i), where b is a constant.
It is given that this plane passes through (0, 3, 0). So,
3 = b
Substituting this value in (i), we get
y = 3, which is the required equation of the plane.

View full question & answer→

Question 553 Marks

Find the equation of a plane which meets the axes at A, B and C, given that the centroid of the triangle ABC is the point $(\alpha,\beta,\gamma)$

Answer

Let a, b and c be the intercepts of the given plane on the coordinate axes.
Then the plane meets the coordinate axes at
A(a, 0, 0), B(0, b, c) and C(0, 0, c)
Given that the centroid of the triangle $=(\alpha,\beta,\gamma)$
$\Rightarrow\Big(\frac{\text{a}+0+0}{3},\frac{0+\text{b}+0}{3},\frac{0+0+\text{c}}{3}\Big)=(\alpha,\beta,\gamma)$
$\Rightarrow\Big(\frac{\text{a}}{3},\frac{\text{b}}{3},\frac{\text{c}}{3}\Big)=(\alpha,\beta,\gamma)$
$\Rightarrow\frac{\text{a}}{3}=\alpha,\frac{\text{b}}{3}=\beta,\frac{\text{c}}{3}=\gamma$
$\Rightarrow\text{a}=3\alpha,\text{b}=3\beta,\text{c}=3\gamma\ ...(\text{i})$
The equation of the plane whose intercepts on the coordinate axes are a, b and c are
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
$\Rightarrow\frac{\text{x}}{3\alpha}+\frac{\text{y}}{3\beta}+\frac{\text{z}}{3\gamma}=1$ [From (i)]
$\Rightarrow\frac{\text{x}}{\alpha}+\frac{\text{y}}{\beta}+\frac{\text{z}}{\gamma}=3$

View full question & answer→

Question 563 Marks

Write the plane $\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})=14$ in normal form.

Answer

The given equation of the plane is
$\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})=14$ or $\vec{\text{r}}.\vec{\text{n}}=14$, where $\vec{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{4+9+36}=7$
For reducing the given equation to normal form, we need to divide it by $|\vec{\text{n}}|$.
Then, we get $\vec{\text{r}}.\frac{\vec{\text{n}}}{|\vec{\text{n}}|}=\frac{14}{|\vec{\text{n}}|}$
$=\vec{\text{r}}.\Big(\frac{2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}}{7}\Big)=\frac{14}{7}$
$\Rightarrow\vec{\text{r}}.\Big(\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=2,$ which is the required normal form.

View full question & answer→

Question 573 Marks

Show that the normals to the following parirs of planes are perpendicular other.
x - y + z - 2 = 0 and 3x + 2y - z + 4 = 0

Answer

Let $\vec{\text{n}_1}$ and $\vec{\text{n}_2}$ be the vectors which are normals to the planes x - y + z = 2 and 3x + 2y - z = -4 respectively.
The given equations of the planes are
x + y + z = 2
3x + 2y - z = -4
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=8,$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=-4$
$\Rightarrow\vec{\text{n}_1}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\vec{\text{n}_2}=3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Now, $\vec{\text{n}_1}\cdot\vec{\text{n}_2}=(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\cdot(3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$=3-2-1=0$
So, the normals to the given planes are perpendicular to each other.

View full question & answer→

Question 583 Marks

Reduce the equation of the following planes to intercept from and find the intercepts on the coordinate axes:
2x + 3y - z = 6

Answer

The equation of the given plane is,

2x + 3y - z = 6

Dividng both sides by 6, we get

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

$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{2}+\frac{\text{z}}{-6}=1\ ...(\text{i})$

We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is,

$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{ii})$

Comparing (i) and (ii) we get

$\text{a}=3;\text{ b}=2;\text{ c}=-6$

View full question & answer→

Question 593 Marks

Find the equation of the plane passing through the following points:
(2, 1, 0), (3, -2, -2) and (3, 1, 7)

Answer

The equation of the plane passing through points (2, 1, 0), (3, -2, -2) and (3, 1, 7) is given by,

$\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0\\3-2&-2-1&-2-0\\3-2&1-1&7-0\end{vmatrix}=0$

$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0\\1&-3&-2\\1&0&7\end{vmatrix}=0$

$\Rightarrow-21(\text{x}-2)-9(\text{y}-1)+3\text{z}=0$

$\Rightarrow-21\text{x}+42-9\text{y}+9+3\text{z}=0$

$\Rightarrow-21\text{x}-9\text{y}+3\text{z}+51=0$

$\Rightarrow21\text{x}+9\text{y}-3\text{z}=51$

$\Rightarrow7\text{x}+3\text{y}-\text{z}=17$

View full question & answer→

Question 603 Marks

Reduce the equation of the following planes to intercept from and find the intercepts on the coordinate axes:
2x - y + z = 5

Answer

Equation of the given plane is,

2x - y + z = 5

Dividng both sides by 5, we get

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

$\Rightarrow\frac{\text{x}}{\big(\frac{5}{2}\big)}+\frac{\text{y}}{-5}+\frac{\text{z}}{5}=1\ ...(\text{i})$

We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is,

$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{ii})$

Comparing (i) and (ii) we get

$\text{a}=\frac{5}{2};\text{ b}=-5;\text{ c}=5$

View full question & answer→

Question 613 Marks

Show that the normals to the following parirs of planes are perpendicular other.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})=5$

Answer

Let $\vec{\text{n}_1}$ and $\vec{\text{n}_2}$ be the vectors which are normals to the planes $\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})=5$ respectively.

The given equations of the planes are

$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})=5$

$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})=5$

$\Rightarrow\vec{\text{n}_1}=(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}),$

$\Rightarrow\vec{\text{n}_2}=(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})$

Now, $\vec{\text{n}_1}\cdot\vec{\text{n}_2}=(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})\cdot(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})$

$=4+2-6=0$

So, the normals to the given planes are perpendicular to each other.

View full question & answer→

Maths 3 Marks