True False[1 Marks ]

Question 11 Mark

State True or False for the following:
If the foot of perpendicular drawn from the origin to a plane is (5, -3, -2), then the equation of plane is $\vec{\text{r}}\cdot(5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}})=38.$

Answer

True.Solution:
We are given that, the required plane passes through the point P(5, -3, -2) and is perpendicular to $\overrightarrow{\text{OP}}$
$\therefore\vec{\text{a}}=5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{n}}=\overrightarrow{\text{OP}}=5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}$
Now, the equation of the plane is
$(\vec{\text{r}}-\vec{\text{a}})\cdot\vec{\text{n}}=0$
$\Rightarrow\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}})$
$=(5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}})\cdot(5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}})$
$=25+9+4$
$\Rightarrow\vec{\text{r}}\cdot(5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}})=38$

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Question 21 Mark

State True or False for the following:
The angle between the planes $\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+\text{k})=1$ and $\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}})=4$ is $\cos^{-1}\Big(\frac{-5}{\sqrt{58}}\Big).$

Answer

False.Solution:
The angle between two planes $\vec{\text{r}}.\vec{\text{n}}_1=\text{d}_1$ and $\vec{\text{r}}.\vec{\text{n}}_2=\text{d}_2$ is given by $\cos\theta=\frac{|\vec{\text{n}}_1\cdot\vec{\text{n}}_2|}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}$
Comparing $\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+\text{k})=1$ and $\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}})=4$ with $\vec{\text{r}}.\vec{\text{n}}_1=\text{d}_1$ and $\vec{\text{r}}.\vec{\text{n}}_2=\text{d}_2,$ we get
$\vec{\text{n}}_1=(2\hat{\text{i}}-3\hat{\text{j}}+\text{k})$ and $\vec{\text{n}}_2=(\hat{\text{i}}-\hat{\text{j}})$
Substituting $\vec{\text{n}}_1=(2\hat{\text{i}}-3\hat{\text{j}}+\text{k})$ and $\vec{\text{n}}_2=(\hat{\text{i}}-\hat{\text{j}})$ in $\cos\theta=\frac{|\vec{\text{n}}_1\cdot\vec{\text{n}}_2|}{|\vec{\text{n}}_1||\vec{\text{n}}_2|},$
$\cos\theta=\frac{\big|(2\hat{\text{i}}-3\hat{\text{j}}+\text{k})(\hat{\text{i}}-\hat{\text{j}})\big|}{\sqrt{4+9+1\sqrt{1+1}}}$
$\Rightarrow\cos\theta=\frac{|2+3|}{\sqrt{14}\cdot\sqrt{2}}=\frac{5}{2\sqrt{7}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{5}{2\sqrt{7}}\Big)$

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Question 31 Mark

State True or False for the following:
The unit vector normal to the plane x + 2y +3z – 6 = 0 is $\frac{1}{\sqrt{14}}\hat{\text{i}}+\frac{2}{\sqrt{14}}\hat{\text{j}}+\frac{3}{\sqrt{14}}\hat{\text{k}}.$

Answer

True.Solution:
The unit vector normal to the plane ax + by + cz + d = 0 is given by
$\vec{\text{n}}=\frac{\text{a}\hat{\text{i}}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}+\frac{\text{b}\hat{\text{j}}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}+\frac{\text{c}\hat{\text{k}}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
Comparing x + 2y + 3z - 6 = 0 with ax + by + cz + d = 0, we get a = 1, b = 2 and c = 3
Substituting a = 1, b = 2 and c = 3 in $\vec{\text{n}}=\frac{\text{a}\hat{\text{i}}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}+\frac{\text{b}\hat{\text{j}}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}+\frac{\text{c}\hat{\text{k}}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},$ we get
$\vec{\text{n}}=\frac{\text{a}\hat{\text{i}}}{\sqrt{1^2+2^2+3^2}}+\frac{\text{b}\hat{\text{j}}}{\sqrt{1^2+2^2+3^2}}+\frac{\text{c}\hat{\text{k}}}{\sqrt{1^2+2^2+3^2}}$
$\vec{\text{n}}=\frac{\hat{\text{i}}}{\sqrt{14}}+\frac{2\hat{\text{j}}}{\sqrt{14}}+\frac{\hat{3\text{k}}}{\sqrt{14}}$

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Question 41 Mark

State True or False for the following:
The line $\vec{\text{r}}=2\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}+\lambda({\text{i}}-{\text{j}}+2{\text{k}})$ lies in the plane $\vec{\text{r}}\cdot(3\hat{\text{i}}+{\text{j}}-{\text{k}})+2=0$

Answer

False:Solution:
We have, $\vec{\text{r}}=2\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}+\lambda({\text{i}}-{\text{j}}+2{\text{k}})$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})=(2+\lambda)\hat{\text{i}}+(-3-\lambda)\hat{\text{j}}+(-1+2\lambda)\hat{\text{k}}$
Position vector of any point on this line is
$(2+\lambda)\hat{\text{i}}+(-3-\lambda)\hat{\text{j}}+(-1+2\lambda)\hat{\text{k}}$
if this point lies on the then L.H.S of the plane is
$-6+3\lambda-3-\lambda+1-2\lambda+2$
$\neq0$
So, the line does not lie on the plane.

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Question 51 Mark

State True or False for the following:
The angle between the line $\vec{\text{r}}=(5\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}})+\lambda(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}})$ and the plane $\vec{\text{r}}(3\hat{\text{i}}-4\hat{\text{j}}-\hat{\text{k}})+5=0$ is $\sin^{-1}\Big(\frac{5}{2\sqrt{91}}\Big).$

Answer

False.Solution:
We have, $\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{n}}=3\hat{\text{i}}-4\hat{\text{j}}-\hat{\text{k}}$
Let $\theta$ is the angle between line and plane.
Then, $\sin\theta=\frac{|\vec{\text{b}}\cdot\vec{\text{n}}|}{|\vec{\text{b}}|\cdot|\vec{\text{n}}|}$
$=\bigg|\frac{(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k})}\cdot(3\hat{\text{i}}-4\hat{\text{j}}-\hat{\text{k}})}{\sqrt{6}\cdot\sqrt{26}}\bigg|$
$=\frac{|6+4-1|}{\sqrt{156}}=\frac{9}{2\sqrt{39}}$
$\therefore\theta=\sin^{-1}\frac{9}{2\sqrt{39}}$

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Question 61 Mark

State True or False for the following:
The equation of a line, which is parallel to $2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$ and which passes through the point (5, -2, 4) is $\frac{\text{x}-5}{2}=\frac{\text{y}-5}{-1}=\frac{\text{z}-4}{3}.$

Answer

False.Solution:
Line is parallel to the vector $2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}$
Line passing throught the point (5, -2, 4)
So its equation is $\frac{\text{x}-5}{2}=\frac{\text{y}-5}{-1}=\frac{\text{z}-4}{3}$

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Question 71 Mark

State True or False for the following:
The vector equation of the line $\frac{\text{x}-5}{3}=\frac{\text{y}-4}{7}=\frac{\text{z}-6}{2}$ is $\vec{\text{r}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}+\lambda(3\hat{\text{i}}-7\hat{\text{j}}+2\hat{\text{k}}).$

Answer

True.Solution:
We have x = 5, y = -4, z = 6
and a = 3, b = 7, c = 2
$\therefore\vec{\text{r}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}+\lambda(3\hat{\text{i}}-7\hat{\text{j}}+2\hat{\text{k}})$

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Question 81 Mark

State True or False for the following:
The intercepts made by the plane 2x - 3y + 5z + 4 = 0 on the co-ordinate axis are $-2, \frac{4}{3},-\frac{4}{5}.$

Answer

True.Solution:
We have equation of plane as $2\text{x}-3\text{y}+5\text{z}+4=0$
$\Rightarrow2\text{x}-3\text{y}+5\text{z}=-4$
$\Rightarrow\frac{2\text{x}}{-4}-\frac{3\text{y}}{-4}+\frac{5\text{z}}{-4}=1$
$\Rightarrow\frac{\text{x}}{-2}+\frac{\text{y}}{\frac{4}{3}}+\frac{\text{z}}{\big(-\frac{4}{5}\big)}=1$
So, the intercepts $-2,\frac{4}{3}$ and $-\frac{4}{3}.$

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