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MATHS 3 Marks Question

The plane · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 61 questions.

Questions · Page 2 of 2

3 Marks Question

Question 513 Marks
Write the equation of the plane whose intercepts on the coordinate axes are 2, -3 and 4
Answer
Given, intercepts on the coordinate axes are 2, -3 and 4
We know that,
The equation of a plane whose intercept onb the coordinate axes are a, b and c
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{i})$
Here, a = 2, b = -3, c = 4
So,
Equation of required plane is,
$\frac{\text{x}}{2}+\frac{\text{y}}{-3}+\frac{\text{z}}{4}=1$
$\frac{6\text{x}-4\text{y}+3\text{z}}{12}=1$
${6\text{x}-4\text{y}+3\text{z}=}{12}$
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Question 523 Marks
A plane meets the coordinate axes at A, B and C, respectively, such that the centriod of triangle ABC is (1, -2, 3). Find the equation of the plane.
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, 0) and C(0, 0, c)
Given that the centroid of the triangle is (1, -2, 3)
$\Rightarrow\Big(\frac{\text{a}+0+0}{3},\frac{0+\text{b}+0}{3},\frac{0+0+\text{c}}{3}\Big)=(1,-2,3)$
$\Rightarrow\Big(\frac{\text{a}}{3},\frac{\text{b}}{3},\frac{\text{c}}{3}\Big)=(1,-2,3)$
$\Rightarrow\frac{\text{a}}{3}=1,\frac{\text{b}}{3}=-2,\frac{\text{c}}{3}=3$
$\Rightarrow\text{a}=3,\text{b}=-6,\text{c}=9\ ...(\text{i})$
Equation of the plane whose intercepts on the coordinate axes a, b and c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{-6}+\frac{\text{z}}{9}=1$ [From (i)]
$\Rightarrow6\text{x}-3\text{y}+2\text{x}=18$
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Question 533 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.
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Question 543 Marks
Show that the following point are coplanar:
(0, 4, 3), (-1, -5, -3), (-2, -2, 1) and (1, 1, -1)
Answer
The equation of the plane passing through points (0, 4, 3), (-1, -5, -3), (-2, -2, 1) is, $\begin{vmatrix}\text{x}-0&\text{y}-4&\text{z}-3\\-1-0&-5-4&-3-3\\-2-0&-2-4&1-3\end{vmatrix}=0$ $\Rightarrow\begin{vmatrix}\text{x}&\text{y}-4&\text{z}-3\\-1&-9&-6\\-2&-6&-2\end{vmatrix}=0$ $\Rightarrow-18\text{x}+10(\text{y}-4)-12(\text{z}-3)=0$ $\Rightarrow9\text{x}-5(\text{y}-4)+6(\text{z}-3)=0$ $\Rightarrow9\text{x}-5\text{y}+\text{z}+2=0$ Substituting the last points (1, 1, -1) (it means x = 1; y = 1; z = -1) in this plane equation, we get9(1) - 5(1) + 6(-1) + 2 = 0
⇒ 4 - 4 = 0 ⇒ 0 = 0 So, the plane equation is satisfied by the points (1, 1, -1) So, the given pointsa are coplanar.
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Question 553 Marks
Obtain the equation of the plane passing through the point (1, - 3, -2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.
Answer
The equation of any plane passing through (1, -3, -2) is,
a(x - 1) + b(y + 3) + c(z + 2) = 0 ....(i)
It is given that (i) is perpendicular to each of the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8. Then,
a + 2b + 2c = 0 ....(ii)
3a + 3b + 2c = 0 .....(ii)
Solving (i), (ii) and (iii) we get,
$\begin{vmatrix}\text{x}-1&\text{y}+3&\text{z}+2\\1&2&2\\3&3&2\end{vmatrix}=0$
⇒ -2(x - 1) + 4(y + 3) - 3(z + 2) = 0
⇒ -2x + 2 + 4y + 12 - 3z - 6 = 0
⇒ 2x - 4y + 3z - 8 = 0
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Question 563 Marks
Find the angle between the line $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{-1}=\frac{\text{z}-3}{2}$ and the plane 3x + 4y + z + 5 = 0
Answer
The given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and the given plane id normal to the vector $\vec{\text{n}}=3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}$
We know that the angle $\theta$ between the line and the plane is given by
$\sin\theta=\frac{\vec{\text{b}}\cdot\vec{\text{n}}}{|\vec{\text{b}}|\vec{|\text{n}}|}$
$=\frac{(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})\cdot(3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}})}{|3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}||3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}|}$
$=\frac{9-4+2}{\sqrt{9+1+4}\sqrt{9+16+1}}=\frac{7}{\sqrt{14}\sqrt{26}}$
$=\frac{7}{\sqrt{2}\sqrt{7}\sqrt{2}\sqrt{13}}=\frac{\sqrt{7}}{\sqrt{52}}=\sqrt{\frac{7}{52}}$
$\theta=\sin^{-1}\Big(\sqrt{\frac{7}{52}}\Big)$
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Question 573 Marks
Find the equation of the plane passing through the following point:
(-5, 0, -6), (-3, 10, -9) and (-2, 6, -6)
Answer
The equation of the plane passing through points (-5, 0, -6), (-3, 10, -9) and (-2, 6, -6) is given by,
$\begin{vmatrix}\text{x}+5&\text{y}-0&\text{z}+6\\-3+5&10-0&-9+6\\-2+5&6-0&-6+6\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}+5&\text{y}&\text{z}+6\\2&10&-3\\3&6&0\end{vmatrix}=0$
$\Rightarrow18(\text{x}+5)-\text{y}-18(\text{z}+6)=0$
$\Rightarrow2(\text{x}+5)-\text{y}-2(\text{z}+6)=0$
$\Rightarrow2\text{x}+10-\text{y}-2\text{z}-12=0$
$\Rightarrow2\text{x}-\text{y}-2\text{z}-2=0$
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Question 583 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.
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Question 593 Marks
Show that the following point are coplanar:
(0, -1, 0), (2, 1, -1), (1, 1, 1) and (3, 3, 0)
Answer
The equation of the plane passing through points (0, -1, 0), (2, 1, -1) and (1, 1, 1) is given by, $\begin{vmatrix}\text{x}-0&\text{y}+1&\text{z}-0\\2-0&1+1&-1-0\\1-0&1+1&1-0\end{vmatrix}=0$ $\Rightarrow\begin{vmatrix}\text{x}&\text{y}+1&\text{z}\\2&2&-1\\1&2&1\end{vmatrix}=0$ $\Rightarrow4\text{x}-3(\text{y}+1)+2\text{z}=0$ $\Rightarrow4\text{x}-3\text{y}+2\text{z}-3=0$ Substituting the last points (3, 3, 0) (it means x = 3; y = 3; z = 0) in this plane equation, we get4(3) - 3(3) + 2(0) - 3 = 0
⇒ 12 - 12 = 0 ⇒ 0 = 0 So, the plane equation is satisfied by the points (3, 3, 0) So, the given pointsa are coplanar.
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Question 603 Marks
Show that the following planes are at right angles.
$x - 2y + 4z = 10$ and $18x + 17y + 4z = 49$
Answer
We know that the planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perperndicular to each other only if
$a_1a_2 + b_1b_2 + c_1c_2= 0$
The given planes are $x - 2y + 4z = 10$ and $18x + 17y + 4z = 49$
$\Rightarrow a_1 = 1; b_1 = -2; c_1 = 4; a_2 = 18; b_2 = 17; c_2 = 4$
Now,
$a_1a_2 + b_1b_2 + c_1c_2$
$= (1)(18) + (-2)(17) + (4)(4)$
$= 18 - 34 + 16$
$= 0$
So, the given planes are perpendicular.
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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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