3 Marks Question

Question 513 Marks

Find the vector and cartesian equations of a plane passing through the point (1, -1, 1) and normal to the line joining the points (1, 2, 5) and (-1, 3, 1)

Answer

Since the given plane passes through the point (1, -1, 1) and is normal to the line joining A(1, 2, 5) and B(-1, 3, 1)
$\overrightarrow{\text{n}}=\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(-\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}})-(\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}})$
$=-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$
We know that the vector equation of the passing through a point $\overrightarrow{\text{a}}$ and normal to $\overrightarrow{\text{n}}$ is,
$\overrightarrow{\text{r}}\cdot\overrightarrow{\text{n}}=\overrightarrow{\text{a}}\cdot\overrightarrow{\text{n}}$
Substituting $\overrightarrow{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\overrightarrow{\text{n}}=-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}},$ we get
$\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})$
$\Rightarrow\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=-2-1-4$
$\Rightarrow\overrightarrow{\text{r}}\cdot\big[-(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})\big]=-7$
$\Rightarrow\overrightarrow{\text{r}}\cdot(-2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}})=7$
For cartesian form, we need to substitute $\overrightarrow{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation.
Then, we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})=7$
$\Rightarrow2\text{x}-\text{y}+4\text{z}=7$

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Question 523 Marks

Reduce the equation of the following planes to intercept from and find the intercepts on the coordinate axes:
4x + 3y - 6z - 12 = 0

Answer

Equation of the given plane is,
4x + 3y - 6z - 12 = 0
⇒ 4x + 3y - 6z = 12
Dividng both sides by 12, we get
$\frac{4\text{x}}{12}+\frac{3\text{y}}{12}+\frac{(6\text{z})}{12}=\frac{12}{12}$
$\Rightarrow\frac{4\text{x}}{12}+\frac{3\text{y}}{12}-\frac{6\text{z}}{12}=\frac{12}{12}$
$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{4}+\frac{\text{z}}{-2}=1\ ...(\text{i})$
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
$\text{a}=3;\text{ b}=4;\text{ c}=-2$

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Question 533 Marks

Find the distance of the point P(-1, -5, -10) from the point of intersection of the line joining the points A(2, -1, 2) and B(5, 3, 4) with the plane x - y + z = 5.

Answer

Equation of the line through the points A(2, -1, 2) and B(5, 3, 4) is $\frac{\text{x}-2}{5-2}=\frac{\text{y}+1}{3+1}=\frac{\text{z}-2}{4-2}=\text{r}$
$\Rightarrow\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-2}{2}=\text{r}$
$\Rightarrow\text{x}=3\text{r}+2,\text{ y}=4\text{r}-1,\text{ z}=2\text{r}+2$
Substituting these in the plane equation we get
$(3\text{r}+2)-(4\text{r}-1)+(2\text{r}+2)=5$
$\Rightarrow\text{r}=0$
$\Rightarrow\text{x}=2,\text{ y}=-1,\text{ z}=2$
Distance of (2, -1, 2) from (-1, -5, -10) is
$=\sqrt{(2-(-1))^2+(-1-(-5))^2+(2-(-10))^2}=\sqrt{3^2+4^2+12^2}$
$=\sqrt{169}=13$

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Question 543 Marks

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the:
zx-plane

Answer

Direction ratios of the given line are
(5 - 3, 1 - 4, 6 - 1) = (2, -3, 5)
Hence, equation of the line is
$\frac{\text{x}-5}{2}=\frac{\text{y}-1}{-3}=\frac{\text{z}-6}{5}=\text{r}$
$\Rightarrow\text{x}=2\text{r}+5,\text{ y}=-3\text{r}+1,\text{ z}=5\text{r}+6$
For any point on the zx-plane y = 0
$\Rightarrow-3\text{r}+1=0$
$\Rightarrow\text{r}=\frac{1}{3}$
$\Rightarrow\text{x}=2\Big(\frac{1}{3}\Big)+5=\frac{17}{3}$
$\Rightarrow\text{z}=5\Big(\frac{1}{3}\Big)+6=\frac{23}{3}$
Hence, the corrdinates of the point are $\Big(\frac{17}{3},0,\frac{23}{3}\Big)$

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Question 553 Marks

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the:
yz-plane

Answer

Direction ratios of the given line are
(5 - 3, 1 - 4, 6 - 1) = (2, -3, 5)
Hence, equation of the line is
$\frac{\text{x}-5}{2}=\frac{\text{y}-1}{-3}=\frac{\text{z}-6}{5}=\text{r}$
$\Rightarrow\text{x}=2\text{r}+5,\text{ y}=-3\text{r}+1,\text{ z}=5\text{r}+6$
For any point on the yz-plane x = 0
$\Rightarrow2\text{r}+5=0$
$\Rightarrow\text{r}=-\frac{5}{2}$
$\Rightarrow\text{y}=-3\Big(-\frac{5}{2}\Big)+1=\frac{17}{2}$
$\Rightarrow\text{z}=5\Big(-\frac{5}{2}\Big)+6=-\frac{13}{2}$
Hence, the corrdinates of the point are $\Big(0,\frac{17}{2},-\frac{13}{2}\Big)$

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Question 563 Marks

Write the vector equation of the line passing through the point (1, -2, -3) and normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=5.$

Answer

The required line is normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=5$ and it is parallel to the normal vector of the plane.
So, the required line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
It is given that the line passes through the point (1, -2, -3) whose position vector is given by $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
We know that the equation of the line passing through the point whose position vector is $\vec{\text{a}}$ and parrallel to the vector $\vec{\text{b}}$ is given by
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\vec{\text{r}}=\big(\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$

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Question 573 Marks

Find the angle between the plane:
2x + y - 2z = 5 and 3x - 6y - 2z = 7

Answer

We know that the angle between the planes,
$\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1=0$ and $\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2=0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}^2_2}}$
So, the angle between 2x + y - 2z = 5 and 3x - 6y - 2z = 7 is given by
$\cos\theta=\frac{(2)(3)+(1)(-6)+(-2)(-2)}{\sqrt{2^2+1^2+(-2)^2}\sqrt{3^2+(-6)^2+(-2)^2}}$
$=\frac{6-6+4}{\sqrt{4+1+4}\sqrt{9+36+4}}$
$=\frac{4}{(3)(7)}=\frac{4}{21}$
$\theta=\cos^{-1}\Big(\frac{4}{21}\Big)$

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

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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,$\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0\\3-2&-2-1&-2-0\\3-2&1-1&7-0\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0\\1&-3&-2\\1&0&7\end{vmatrix}=0$
$\Rightarrow-21(\text{x}-2)-9(\text{y}-1)+3\text{z}=0$
$\Rightarrow-21\text{x}+42-9\text{y}+9+3\text{z}=0$
$\Rightarrow-21\text{x}-9\text{y}+3\text{z}+51=0$
$\Rightarrow21\text{x}+9\text{y}-3\text{z}=51$
$\Rightarrow7\text{x}+3\text{y}-\text{z}=17$

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

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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.

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MATHS 3 Marks Question