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Maths 4 Marks

Three Dimensional Geometry · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 343 questions.

Questions · Page 2 of 7

4 Marks

Question 514 Marks
Find the angle between the following pairs of lines:

$\frac{\text{x}-1}{2}=\frac{\text{y}-2}{3}=\frac{\text{z}-3}{-3}$ and $\frac{\text{x}+3}{-1}=\frac{\text{y}-5}{8}=\frac{\text{z}-1}{4}$

Answer
$\frac{\text{x}-1}{2}=\frac{\text{y}-2}{3}=\frac{\text{z}-3}{-3}$ and $\frac{\text{x}+3}{-1}=\frac{\text{y}-5}{8}=\frac{\text{z}-1}{4}$
Let $\vec{\text{b}}_1$ and $\vec{\text{b}}_2$ be vectors parallel to the given lines.
Now,
$\vec{\text{b}}_1=2\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{b}}_2=-\hat{\text{i}}+8\hat{\text{j}}+4\hat{\text{k}}$
If $\theta$ is the angle between the given lines, then
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(2\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}\big).\big(-\hat{\text{i}}+8\hat{\text{j}}+4\hat{\text{k}}\big)}{\sqrt{2^2+3^2+(-3)^2}\sqrt{(-1)^2+8^2+4^2}}$
$=\frac{-2+24-12}{9\sqrt{22}}$
$=\frac{10}{9\sqrt{22}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{10}{9\sqrt{22}}\Big)$
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Question 524 Marks
Prove that the lines through A(0, -1, -1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(-4, 4, 4). Also, find their point of intersection.
Answer
The coordinates of any point on the line AB are given by
$\frac{\text{x}-0}{4-0}=\frac{\text{y}+1}{5+1}=\frac{\text{z}+1}{1+1}=\lambda$
$\Rightarrow\text{x}=4\lambda$
$\text{y}=6\lambda-1$
$\text{z}=2\lambda-1$
The coordinates of a general point on AB are $4\lambda,6\lambda-1,2\lambda-1.$
The coordinates of any point on the line CD are given by
$\frac{\text{x}-3}{3+4}=\frac{\text{y}-9}{9-4}=\frac{\text{z}-4}{4-4}=\mu$
$\Rightarrow\text{x}=7\mu+3$
$\text{y}=5\mu+9$
$\text{z}=4$
The coordinates of a general point on CD are $7\mu+3,5\mu+9,4.$
If the lines AB and CD intersect, then they have a common point. so, for some valuse of $\lambda$ and $\mu,$
We must have
$4\lambda=7\mu+3,6\lambda-1=5\mu+9,2\lambda-1=4$
$\Rightarrow4\lambda-7\mu=3\dots(1)$
$6\lambda-5\mu=10\dots(2)$
$\lambda=\frac{5}{2}\dots(3)$
Solving (2) and (3), we get
$\lambda=\frac{5}{2}$
$\mu=1$
Substituting $\lambda=\frac{5}{2}$ and $\mu=1$ in (1), we get
$\text{LHS}=4\lambda-7\mu$
$=4\Big(\frac{5}{2}\Big)-7(1)$
$=3$
$=\text{RHS}$
Since $\lambda=\frac{5}{2}$ and $\mu=1$ satisfy (3), the given lines intersect.
substituting the value of $\lambda$ in the coordinates of a general point on the line AB, we get
x = 10
y = 14
z = 4
Hence, AB and CD intersect at point (10, 14, 4).
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Question 534 Marks
Find the distance between the lines l1 and l2 given by

$\vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$ and, $\vec{\text{r}}=3\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}+\mu\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$

Answer
$\vec{\text{a}}_1=\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{a}}_2=3\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$
$\vec{\text{a}}_2-\vec{\text{a}}_1=3\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$
$=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&1&-1\\2&3&6\end{vmatrix}$
$=\hat{\text{i}}(6+3)-\hat{\text{j}}(12+2)+\hat{\text{k}}(6-2)$
$=9\hat{\text{i}}-14\hat{\text{j}}+4\hat{\text{k}}$
Shortest distance between 2 lines
$=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}}{\big|\vec{\text{b}}\big|}\Bigg|$
$=\Bigg|\frac{9\hat{\text{i}}-14\hat{\text{j}}+4\hat{\text{k}}}{\big|\sqrt{2^2+3^2+6^2}\big|}\Bigg|$
$=\Bigg|\frac{9\hat{\text{i}}-14\hat{\text{j}}+4\hat{\text{k}}}{\sqrt{49}}\Bigg|$
$=\Bigg|\frac{\sqrt{9^2+(-14)^2+4^2}}{\sqrt{49}}\Bigg|$
$=\Big|\frac{\sqrt{293}}{\sqrt{49}}\Big|=\frac{\sqrt{293}}{7}\text{ units}$
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Question 544 Marks
Find the angle between the follwing pairs of lines:

$\vec{\text{r}}=\big(3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(5\hat{\text{j}}-2\hat{\text{k}}\big)+\mu\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$

Answer
$\vec{\text{r}}=\big(3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(5\hat{\text{j}}-2\hat{\text{k}}\big)+\mu\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$
Let b1 and b2 be vectors parallel to the given lines.
Now,
$\vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}_2=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
If $\theta$ is the angle between the given lines, then
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big).\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)}{\sqrt{1^1+2^2+2^2}\sqrt{3^2+2^2+6^2}}$
$=\frac{3+4+12}{3\times7}$
$=\frac{19}{21}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{19}{21}\Big)$
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Question 554 Marks
Find the vector equation of the line passing through the points (-1, 0, 2) and (3, 4, 6).
Answer
We know that the vector equation of a line passing through the points with position vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\big(\vec{\text{b}}-\vec{\text{a}}\big) $ where $\lambda$ is a scalar.
Here,
$\vec{\text{a}}=-1\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}}$
Vector equation of the required line is
$\vec{\text{r}}=\big(-1\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}}\big)+\lambda\big\{\big(3\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}}\big)-\big(-1\hat{\text{i}}+0\hat{\text{j}}+12\hat{\text{k}}\big)\big\}$
$\Rightarrow\vec{\text{r}}=\big(-1\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}}\big)+\lambda\big(4\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}\big)$
Here, $\lambda$ is a parameter.
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Question 564 Marks
Find the angle between the following pairs of lines:

$\frac{\text{x}-2}{3}=\frac{\text{y}+3}{-2},\text{z}=5$ and $\frac{\text{x}+1}{1}=\frac{2\text{y}-3}{3}=\frac{\text{z}-5}{2}$

Answer
$\frac{\text{x}-2}{3}=\frac{\text{y}+3}{-2},\text{z}=5$ and $\frac{\text{x}+1}{1}=\frac{2\text{y}-3}{3}=\frac{\text{z}-5}{2}$

The equations of the given lines can be re-written as

$\frac{\text{x}-2}{3}=\frac{\text{y}+3}{-2}=\frac{\text{z}-5}{0}$ and $\frac{\text{x}+1}{1}=\frac{\text{y}-\frac{3}{2}}{\frac{3}{2}}=\frac{\text{z}-5}{2}$

Let $\vec{\text{b}}_1$ and $\vec{\text{b}}_2$ be vectors parallel to the given lines.

Now,

$\vec{\text{b}}_1=3\hat{\text{i}}-2\hat{\text{j}}+0\hat{\text{k}}$

$\vec{\text{b}}_2=\hat{\text{i}}+\frac{3}{2}\hat{\text{j}}+2\hat{\text{k}}$

If $\theta$ is the angle between the given lines, then

$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$

$=\frac{\big(3\hat{\text{i}}-2\hat{\text{j}}+0\hat{\text{k}}\big).\big(\hat{\text{i}}+\frac{3}{2}\hat{\text{j}}+2\hat{\text{k}}\big)}{\sqrt{3^2+(-2)^2+0^2}\sqrt{1^2+\Big(\frac{3}{2}\Big)+2^2}}$

$=\frac{3-3+0}{\sqrt{13}\sqrt{\frac{29}{4}}}$

$=0$

$\Rightarrow\theta=\frac{\pi}{2}$

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Question 574 Marks
Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line $\frac{-\text{x}-2}{1}=\frac{\text{y}+3}{7}=\frac{2\text{z}-6}{3}.$
Answer
we know that, equation of a line passing through a point (x1, y1, z1) and having direction ratios proportional to a, b, c is
$\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}\dots(1)$
Here, (x1, y1, z1) = (1, 2, 3) and 
Given line $\frac{-\text{x}-2}{1}=\frac{\text{y}+3}{7}=\frac{2\text{z}-6}{3}$
$\Rightarrow\frac{\text{x}+2}{-1}=\frac{\text{y}+3}{7}=\frac{\text{z}-6}{\frac{3}{2}}$
It parallel to the required line, so
$\text{a}=\mu,\text{b}7\mu,\text{c}=\frac{3}{2}\mu$
So, equation of required line using equation (1) is,
$\frac{\text{x}-1}{-\mu}=\frac{\text{y}-2}{7\mu}=\frac{\text{z}-3}{\frac{3}{2}\mu}$
$\Rightarrow\frac{\text{x}-1}{-1}=\frac{\text{y}-2}{7}=\frac{\text{z}-3}{\frac{3}{2}}$
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Question 584 Marks
Find the cartesian equation of a line passing through (1, -1, 2) and parallel to the line whose equation are $\frac{\text{x}-3}{1}=\frac{\text{y}-1}{2}=\frac{\text{z}+1}{-2}.$ Also, reduce the equation obtained in vector form.
Answer
We know that the cartesian equation of a line passing through a point with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{m}}$ is $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}.$
Here,
$\vec{\text{a}}=\text{x}_1\hat{\text{i}}+\text{y}_1\hat{\text{j}}+\text{z}_1\hat{\text{k}}$
$\vec{\text{m}}=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$
Here, $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
Cartesian equation of the required line is 
$\frac{\text{x}-1}{1}=\frac{\text{y}-(-1)}{2}=\frac{\text{z}-2}{-2}$
$\Rightarrow\frac{\text{x}-1}{1}=\frac{\text{y}+1}{2}=\frac{\text{z}-2}{-2}$
We know that the cartesian equation of a line passing through a points with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{m}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{m}}.$
Here, the line is passing through the point (1, 1, -2) and its direction ration are proportional to 1, 2, -2.
Vector equation of the required line is
$\vec{\text{r}}=\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\big)$
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Question 594 Marks
Find the foot of the perpendicular from (0, 2, 7) on the line $\frac{\text{x}+2}{-1}=\frac{\text{y}-1}{3}=\frac{\text{z}-3}{-2}.$
Answer
Let L be the foot of the perpendicular drawn from the point P (0, 2, 7) to the given line.

The coordoinates of a general point on the line $\frac{\text{x}+2}{-1}=\frac{\text{y}-1}{3}=\frac{\text{z}-3}{-2}$ are given by

$\frac{\text{x}+2}{-1}=\frac{\text{y}-1}{3}=\frac{\text{z}-3}{-2}=\lambda$

$\Rightarrow\text{x}=-\lambda-2$

$\text{y}=3\lambda+1$

$\text{z}=-2\lambda+3$

Let the coordinates of L be $-\lambda-2,3\lambda+1,-2\lambda+3.$

The direction ratios of PL are proportional to $-\lambda-2-0,3\lambda+1-2,-2\lambda+3-7,$

i,e. $-\lambda-2,3\lambda-1,-2\lambda-4.$

The direction ratios of the given line are proportionl to -1, 3, -2, but PL is perpendicular to the given line.

$\therefore-1-\lambda-2+33\lambda-1-2\lambda-4=0\Rightarrow\lambda=-12$

Substituting $\lambda=-12$ in $-\lambda-2,3\lambda+1.-2\lambda+3,$ we get the coordinates of L as -32, -12, 4.

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Question 604 Marks
ABCD is aparallelogram. the position vectora of the points A, B and C are respectively, $4\hat{\text{i}}+5\hat{\text{j}}-10\hat{\text{k}},2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and $-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}.$ Find the vector equation of the line BD. Also, reduce it to cartesian form.
Answer
We know that the position vector of the mid-point of $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\vec{\text{a}}+\vec{\text{b}}}{2}.$
Let the position of mid-point of A and C= position vector of mid-point of B and D
$\therefore\frac{\big(4\hat{\text{i}}+5\hat{\text{j}}-10\hat{\text{k}}\big)+\big(-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)}{2}=\frac{\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)}{2}$
$\Rightarrow\frac{3}{2}\hat{\text{i}}+\frac{7}{2}\hat{\text{j}}-\frac{9}2{}\hat{\text{k}}=\Big(\frac{\text{x}+2}{2}\Big)\hat{\text{i}}+\Big(\frac{-3+\text{y}}{2}\Big)\hat{\text{j}}+\Big(\frac{4+\text{z}}{2}\Big)\hat{\text{k}}$
Comparing the coeffient of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$\frac{\text{x}+2}{2}=\frac{3}{2}$
$\Rightarrow\text{x}=1$
$\frac{-3+\text{y}}{2}=\frac{7}{2}$
$\Rightarrow\text{y}=10$
$\frac{4+\text{z}}{2}=-\frac{9}{2}$
$\Rightarrow\text{z}=-13$
Position vector of point $\text{D}=\hat{\text{i}}+10\hat{\text{j}}-13\hat{\text{k}}$
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Question 614 Marks
Find the angle between the following pairs of lines:

$\frac{\text{x}-5}{1}=\frac{2\text{y}+6}{-2}=\frac{\text{z}-3}{1}$ and $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-6}{5}$

Answer
$\frac{\text{x}-5}{1}=\frac{2\text{y}+6}{-2}=\frac{\text{z}-3}{1}$ and $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-6}{5}$

The equation of the given lines can be re-written as

$\frac{\text{x}-5}{1}=\frac{\text{y}+3}{-1}=\frac{\text{z}-3}{1}$ and $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{4}=\frac{\text{z}-6}{5}$

Let $\vec{\text{b}}_1$ and $\vec{\text{b}}_2$ be vectors parallel to the given lines.

Now,

$\vec{\text{b}}_1=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$

$\vec{\text{b}}_2=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$

If $\theta$ is the angle between the given lines, then

$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$

$=\frac{\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big).\big(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)}{\sqrt{1^2+(-1)^2+1^2}\sqrt{3^2+4^2+5^2}}$

$=\frac{3-4+5}{\sqrt{3}\sqrt{50}}$

$=\frac{4}{5\sqrt{6}}$

$\Rightarrow\theta=\cos^{-1}\Big(\frac{4}{5\sqrt{6}}\Big)$

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Question 624 Marks
If the coordinates of the points A, B, C, are (1, 2, 3), (4, 5, 6), (-4, 3, -6) and (2, 9, 2), then find the angle between AB and CD.
Answer
The given points are A(1, 2, 3), B(4, 5, 6), C(-4, 3, -6) and D(2, 9, 2).

We know that the direction ratios of the line joining the points

(x1, y1, z1) and (x2, y2, z2) are x2 x1, y2 y1, z2 z1. 2

The direction ratios of AB are (4 - 1), (5 - 2), (7 - 3),i.e. (3, 3, 4).

The direction ratios of CD are [2(-4)],  (9 - 3), [2(-6)],

i.e. 6, 6, 8.

Let, $\theta$ be the angle between AB and CD.

We have, 

$\text{a}_1=3, \text{c}_1=3, \text{c}_1=4$

$\text{a}_2=6, \text{c}_2=6, \text{c}_2=8$

$\therefore\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}_2^2}}$

$=\frac{18+18+32}{\sqrt{9+9+16}\sqrt{36+36+64}}=\frac{68}{68}$

$=1$

$\Rightarrow\theta=0^\circ$

Thus, the angle between AB and CD measures $0^\circ$.

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Question 634 Marks
Find the vector equation of a line which is parallel to the vector $2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and which passes through the point (5, -2, 4). Also, reduce it to cartesian from.
Answer
We know that, vector equation of line passing through a fixed point $\vec{\text{a}}$ and paralel to vector $\vec{\text{b}}$ is

$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}},$ where $\lambda$ is scalar

Here, $\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{a}}=5\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$

So, rquation of required line is

$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$

$\vec{\text{r}}=\big(5\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)$

Put $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}},$ so

$\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)=(5+2\lambda)\hat{\text{i}}+(-2-\lambda)\hat{\text{j}}+(4+3\lambda)\hat{\text{k}}$

Comparing the cofficients of $\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}$ so

$\text{x}=5+2\lambda,\text{y}=-2-\lambda,\text{z}=4+3\lambda$

$\Rightarrow\frac{\text{x}-5}{2}=\lambda,\frac{\text{y}+2}{-0}=\lambda,\frac{\text{z}-4}{3}=\lambda$

Cortesian form of equation of the line is,

$\frac{\text{x}-5}{5}=\frac{\text{y}+2}{-0}=\frac{\text{z}-4}{3}$

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Question 644 Marks
The cartesian equation of a line are 3x + 1 = 6y - 2 =1 - z. Find the fixed point through which it passes, its direction ratios and also its vector equation.
Answer
Given equation of line is,
3x + 1 = 6y - 2 = 1 - z
Dividing all by 6,
$\frac{3\text{x}+1}{6}=\frac{6\text{y}-2}{6}=\frac{1-\text{z}}{6}$
$\Rightarrow\frac{3\text{x}}{6}+\frac{1}{6}=\frac{6\text{y}}{6}-\frac{2}{6}=\frac{1}{6}-\frac{\text{z}}{6}$
$\Rightarrow\frac{1}2{}\text{x}+\frac{1}{6}=\text{y}-\frac{1}{3}=-\frac{\text{z}}{6}+\frac{1}{6}$
$\Rightarrow\frac{1}{2}\Big(\text{x}+\frac{1}{3}\Big)=1\Big(\text{y}-\frac{1}{3}\Big)=+\frac{1}{6}(\text{z}-1)$
$\Rightarrow\frac{\text{x}\frac{1}{3}}{2}=\frac{\text{y}-\frac{1}{3}}{1}=\frac{\text{z}-1}{-6}=\lambda\text{ (say)}\dots(1)$ 
Comparing it with equation of line passing through (x1, y1, z1) and direction ratios a, b, c,
$\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$
$\Rightarrow\big(\text{x}_1,\text{y}_1,\text{z}_1\big)=\Big(-\frac{1}{3},\frac{1}{3},1\Big)$
$\text{a}=2,\text{b}=1,-6$
So, direction ratios of the line are = 2, 1, -6
From equation (1),
$\text{x}=\Big(2\lambda-\frac{1}{3}\Big),\text{y}=\Big(\lambda+\frac{1}{3}\Big),\text{z}=(-6\lambda+1)$
So, vector equation of the given line is,
$\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=\Big(2\lambda-\frac{1}{3}\Big)\hat{\text{i}}+\Big(\lambda+\frac{1}{3}\Big)\hat{\text{j}}+(-6\lambda+1)\hat{\text{k}}$
$\vec{\text{r}}=\Big(-\frac{1}{3}\hat{\text{i}}+\frac{1}{3}\hat{\text{j}}+\hat{\text{k}}\Big)+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}-6\hat{\text{k}}\big)$
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Question 654 Marks
Show that the three lines with direction cosines $\frac{12}{13},\frac{-3}{13},\frac{-4}{13},\frac{4}{13},\frac{12}{13},\frac{3}{13},\frac{3}{13},\frac{-4}{13},\frac{12}{13}$ are mutually perpendicular.
Answer
let $\text{l}_1=\frac{12}{13},\text{m}_1=-\frac{3}{13},\text{n}_1=-\frac{4}{13}$
$\text{l}_2=\frac{4}{13},\text{m}_2=\frac{12}{13},\text{n}_2=\frac{3}{13}$
$\text{l}_3=\frac{3}{13},\text{m}_3=-\frac{4}{13},\text{n}_3=\frac{12}{13}$
$\text{l}_1\text{l}_2+\text{m}_1\text{m}_2+\text{n}_1\text{n}_2$
$=\frac{12}{13}\times\frac{4}{13}+\big(-\frac{3}{13}\big)\times\frac{12}{13}+\big(-\frac{4}{13}\big)\times\frac{3}{13}$
$=\frac{48-36-13}{169}=0$
$\text{l}_2\text{l}_3+\text{m}_2\text{m}_3+\text{n}_2\text{n}_3$
$=\frac{4}{13}\times\frac{3}{13}+\frac{12}{13}\times\Big(-\frac{4}{13}\Big)+\frac{3}{13}\times\frac{12}{13}$
$=\frac{12-48+36}{169}=0$
$\text{l}_1\text{l}_3+\text{m}_1\text{m}_3+\text{n}_1\text{n}_3$
$=\frac{12}{13}\times\frac{3}{13}+\Big(-\frac{3}{13}\Big)\times\Big(-\frac{4}{13}\Big)+\Big(-\frac{4}{13}\Big)\times\frac{12}{13}$
$=\frac{36+12-48}{169}=0$
$\therefore$ The lines are mutually perpendicular.
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Question 664 Marks
Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector $\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}.$ Reduce the corresponding equation in cartesian form.
Answer
We know that the vector equation of a line passing through a point with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}.$
Here,
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
vector equation of the required line is
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
Here, $\lambda$ is a parameter.
Reducing (1) to cartesian form, we get
$\text{x}\hat{\text{i}}+\text{y}\hat{\text{i}}+\text{z}\hat{\text{k}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)$ $\big[\text{putting }\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}} \text{ in }(1)\big]$
$\Rightarrow\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=(1+\lambda)\hat{\text{i}}+(2-2\lambda)\hat{\text{j}}+(3+3\lambda)\hat{\text{k}}$
Comparing the cofficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$\text{x}=1+\lambda,\text{y}=2-2\lambda,\text{z}=3+3\lambda$
$\Rightarrow\text{x}-1=\lambda,\frac{\text{y}-2}{-2}=\lambda,\frac{\text{z}-3}{3}=\lambda$
$\Rightarrow\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-2}=\frac{\text{z}-3}{3}=\lambda$
Hence, the cartesian form of (1) is
$\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-2}=\frac{\text{z}-3}{3}$
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Question 674 Marks
Find the length and the foot ofo perpendicular from the point $\Big(1,\frac{3}{2},2\Big)$ to the plane 2x - 2y + 4z + 5 = 0
Answer
Let M be the foot of the perpendicular from $\text{P}\Big(1,\frac{3}{2},2\Big)$ on the plane 2x - 2y + 4z + 5 = 0
Then, PM is the normal to the plane. So, its directions rations are proportional to 2, -2, 4.
Since PM passes through $\text{P}\Big(1,\frac{3}{2},2\Big)$, therefore, its equation is
$\frac{\text{x}-1}{2}=\frac{\text{y}-\frac{3}{2}}{-2}=\frac{\text{z}-2}{4}=\lambda\text{ (say)}$
Let the coordinates of M be $\Big(2\lambda+1,-2\lambda+\frac{3}{2},4\lambda+2\Big).$
Now, M lies on the plane 2x - 2y + 4z + 5 = 0.
$\therefore\ 2(2\lambda+1)-2\Big(-2\lambda+\frac{3}{2}\Big)+4(4\lambda+2)+5=0$
$\Rightarrow 24\lambda+12=0$
$\Rightarrow \lambda=-\frac{1}{2}$
So, the coordinates of M are $\Big(2\times\Big(-\frac{1}{2}\big)+1,-2\times\Big(-\frac{1}{2}\Big)+\frac{3}{2},4\times\Big(-\frac{1}{2}\Big)+2\Big)$ or $\Big(0,\frac{5}{2},0\Big)$
Thus, the coordinates of the foot of the perpendicular are $\Big(0,\frac{5}{2},0\Big).$
Now,
$\text{PM}=\sqrt{(1-0)^2+\Big(\frac{3}{2}-\frac{5}{2}\Big)^2+(2-0)^2}$
$=\sqrt{1+1+4}$
$=\sqrt{6}$
Thus, the length of the perpendicular from the given point to the plane is $\sqrt{6}$ units.
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Question 684 Marks
Write the vector equation of the following lines and hence determine the distance between them $\frac{\text{x}-1}{2}=\frac{\text{y}-2}{3}=\frac{\text{z}+4}{6}$ and $\frac{\text{x}-3}{4}=\frac{\text{y}-3}{6}=\frac{\text{z}+5}{12}$
Answer
We have
$\frac{\text{x}-1}{2}=\frac{\text{y}-2}{3}=\frac{\text{z}+4}{6}$
$\frac{\text{x}-3}{4}=\frac{\text{y}-3}{6}=\frac{\text{z}+5}{12}$
Since the first line passes line passes through the point (1, 2, -4) and has direction ratios proportional to 2, 3, 6, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1\dots(1)$
$\Rightarrow\vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$
Also, the second line passes through the point (3, 3, -5) and has directional to 4, 6, 12.
Its vector equation is
$\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2\dots(2)$
$\Rightarrow\vec{\text{r}}=3\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}+\mu\big(4\hat{\text{i}}+6\hat{\text{j}}+12\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{r}}=3\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}+2\mu\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$
These two lines pass through the points having position vectors
$\vec{\text{a}}_1=\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$ and $\vec{\text{a}}_2=3\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$ and are parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}.$
Now,
$\vec{\text{a}}_2-\vec{\text{a}}_1=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
and
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)\times\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&1&-1\\2&3&6 \end{vmatrix}$
$=9\hat{\text{i}}-14\hat{\text{j}}+4\hat{\text{k}}$
$\Rightarrow\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|=\sqrt{9^2+(-14)^2+4^2}$
$=\sqrt{81+196+16}$
$=\sqrt{293}$
and $\big|\vec{\text{b}}\big|=\sqrt{2^2+3^2+6^2}$
$=\sqrt{4+9+36}$
$=7$
The shortest distance between the two lines is given by
$\frac{\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|}{\big|\vec{\text{b}}\big|}=\frac{\sqrt{293}}{7}\text{ units}$
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Question 694 Marks
Find the equation of the plane mid-parallel to the planes 2x - 2y + z + 3 = 0 and 2x - 2y + z + 9 = 0
Answer
We know that the distance between two planes ax + by + cz = d1 and ax + by + cz = d2 is $\frac{\big|\text{d}_2-\text{d}_1\big|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
The equation of plane thatb is mid-parallel to the planes
2x - 2y + z + 3 = 0 ...(i)
2x - 2y + z + 9 = 0 ....(ii)
is of the form 2x - 2y + z + k = 0 ...(iii)
It meance that the distance between (i) and (iii) = distance between (i) and (ii)
$\Rightarrow\frac{|\text{k}-3|}{\sqrt{4+4+1}}=\frac{|\text{k}-9|}{\sqrt{4+4+1}}$
⇒ |k - 3| = |k - 9|
⇒ k - 3 = k - 9 or k - 3 = -(k - 9)
⇒ 3 = 9 (false); k - 3 = -k + 9
⇒ 2k = 12
⇒ k = 6
Substituting this in (iii) we get 2x - 2y + z + 6 = 0, which is the required equation of the plane.
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Question 704 Marks
Prove that the line through A(0, –1, –1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(–4, 4, 4).
Answer
Equation of line $\vec{\text{AB}}$
$\vec{\text{r}} = (-\hat{\text{j}} - \hat{\text{k}}) + \lambda (4\hat{\text{i}} + 6\hat{\text{j}} + 2\hat{\text{k}})$
Equation of line $\vec{\text{CD}}$
$\vec{\text{r}} = (3\hat{\text{i}} + 9\hat{\text{j}} + 4\hat{\text{k}}) + \mu (-7\hat{\text{i}} - 5\hat{\text{j}})$
$\vec{\text{a}}_{2} - \vec{\text{a}}_{1} = 3\hat{\text{i}} + 10\hat{\text{j}} + 5 \hat{\text{k}}$
$\vec{\text{b}}_{1} \times \vec{\text{b}}_{2} = \begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\ 4 & 6 & 2 \\ -7 & -5 & 0 \end{vmatrix} = 10\hat{\text{i}} - 14\hat{\text{j}} + 22\hat{\text{k}}$
$(\vec{\text{a}}_{2} - \vec{\text{a}}_{1}). (\vec{\text{b}}_{1} \times \vec{\text{b}}_{2}) = 30 – 140 + 110 = 0$
$\Rightarrow$ Lines intersect.
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Question 714 Marks
Find the perpendicular distence of the point (1, 0, 0) from the line $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-3}=\frac{\text{z}+10}{8}.$ Also, find the coordinates of the perpendicular and the equation of the perpendicular.
Answer
Let foot of the perpemdicular drawn from the point P(1, 0, 0) to the line $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-3}=\frac{\text{z}+10}{8}$ is Q. we have to find lengh of PQ.
Q is a genelar point on the line,
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-3}=\frac{\text{z}+10}{8}=\lambda$ (say)
coordinate of Q $=\big(2\lambda+1,-3\lambda-1,8\lambda-10\big)$
Direction ratios line PQ are
$=\big(2\lambda+1-1\big),\big(-3\lambda-1-0\big),\big(8\lambda-10-0\big)$
$=(2\lambda),\big(-3\lambda-1\big),\big(8\lambda-10\big)$
Since, line PQ is perpendicular to the given line, so
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$(2)(2\lambda)+(-3)(-3\lambda-1)+8(8\lambda-10)=0$
$4\lambda+9\lambda+3+64\lambda-80=0$
$77\lambda-77=0$
$\lambda=1$
Therefore, coordinate of Q is $\big(2\lambda+1,-3\lambda-1,8\lambda-10\big)$
$=\big(2(1)+1,-3(1)-1,8(1)-10\big)$
$=(3,-4,-2)$
$\text{PQ}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$
$=\sqrt{(1-3)^2+(0+4)^2+(0+2)^2}$
$=\sqrt{4+16+4}$
$=\sqrt{24}$
$=2\sqrt{6}$
So, foot of perpendicular $=(3,-1,-2)$
length of perpendicular $=2\sqrt{6}\text{ units}$
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Question 724 Marks
Show that the plane whose vector equation is $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=3$ contains the line whose vector equation is $\vec{\text{r}}=\hat{\text{i}}+\hat{\text{j}}+\lambda(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}).$
Answer
The line $\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}})+\lambda(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})\ ...(\text{i})$
Passes through a point whose posotion vector is $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$
If the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=3$ contains the given line, then
It should passes through the point $\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}.$
It should be parallel to the line.
Now, the plane passes through the point $\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}.$
So, the plane vector to the given plane is $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}.$
We observe that
$\vec{\text{b}}\cdot\vec{\text{n}}=(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$=2+2-4=0$
Therefore, the plane is parallel to the line.
Hence, the given plane contains the given line.
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Question 734 Marks
Find the equation of the plane through the points (2, 1, 0), (3, -2, -2) and (3, 1, 7).
Answer
We know that, the equation of a plane passing through three non-collinear points (x1, x1, x1), (x2, x2, x2) and (x3, x3, x3) is
$\begin{vmatrix}\text{x}-\text{x}_1&\text{y}-\text{y}_1&\text{z}-\text{z}_1 \\\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{x}_3-\text{x}_1&\text{y}_3-\text{y}_1&\text{z}_3-\text{z}_1 \end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0 \\3-2&-2-1&-2-0\\3-2&1-1&7-0 \end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z} \\1&-3&-2\\1&0&7 \end{vmatrix}=0$
$\Rightarrow(\text{x}-2)(-21+0)-(\text{y}-1)(7+2)+\text{z}(3)=0$
$\Rightarrow-21\text{x}+42-9\text{y}+9+3\text{z}=0$
$\Rightarrow-21\text{x}-9\text{y}+3\text{z}=-51$
$\therefore7\text{x}+3\text{y}-\text{z}=17$
So, the required equation of plane is $7\text{x}+3\text{y}-\text{z}=17.$
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Question 744 Marks
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.
Answer
Given: A line through the points A(5, 1, 6) and B(3, 4, 1)
$\therefore$ Direction ratios of this line AB are x2 - x1, y2 - y1, z2 - z1
⇒ 3 - 5, 4 - 1, 1 - 6
⇒ -2, 3, -5 = a, b, c
Equation of the line AB is $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$
$\Rightarrow\ \ \frac{\text{x}-5}{-2}=\frac{\text{y}-1}{3}=\frac{\text{z}-6}{-5}\ \ \ .....(\text{i})$
Now we have to find the coordinates of the point where this line AB crosses the ZX-plane
i.e., y = 0 .......(ii)
Putting y = 0 in eq. (i), we get
$\frac{\text{x}-5}{-2}=\frac{-1}{3}=\frac{\text{z}-6}{-5}$
$\Rightarrow\ \ \frac{\text{x}-5}{-2}=\frac{-1}{3}\ \text{and}\ \frac{\text{z}-6}{-5}=\frac{-1}{3}$
⇒ 3x - 15 = 2 and 3z - 18 = 5
⇒ 3x = 17 and 3z = 23
$\Rightarrow\ \ \ \text{x}=\frac{17}{3}\ \text{and}\ \text{z}=\frac{23}{3}$
Thus, required point is $\text{P}\Big(\ \frac{17}{3},\ 0,\frac{23}{3}\Big).$
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Question 754 Marks
Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane $\vec{\text{r}}.\big(\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}\big)+5=0.$
Answer
The Cartesian equation of the given plane is 2x - 2y + 4z + 5 = 0
Let P(x1, y1, z1) be the foot of perpendicular formula (1, 1, 2) to the plane 2x - 2y + 4z + 5 = 0
Direction ratios of the line PQ are proportional to the direction ratios of the given plane
$\frac{\text{x}_1-1}{1}=\frac{\text{y}_1-1}{-2}=\frac{\text{z}_1-2}{4}=\lambda$
$\Rightarrow\ \text{x}_1=2\lambda+1,\text{y}_1=-2\lambda+1,\text{z}_1=4\lambda+2$
$\text{P}(2\lambda+1,-2\lambda+1,4\lambda+2)$ lies on the plane 2x - 2y + 4z + 5 = 0
$\therefore\ 2(2\lambda+1) -2(2\lambda+1)+4(4\lambda+2)+5=0$
$\Rightarrow 4\lambda+2+4\lambda-2+16\lambda+8+5=0$
$\Rightarrow 24\lambda+13=0$
$\Rightarrow\lambda=-\frac{13}{24}$
$\therefore\ \text{x}_1=2\Big(\frac{-13}{24}\Big)+1=-\frac{-13}{12}+\frac{-1}{12}$
$\text{y}_1=-2\Big(\frac{-13}{24}\Big)+1=\frac{13}{12}+1=\frac{25}{12}$
$\text{z}_1=4\lambda+2=4\Big(\frac{-13}{24}\Big)+4=\frac{-7}{6}$
$\therefore$ Coordinates of foot of perpendicular are $\Big(\frac{-1}{12},\frac{25}{12},\frac{-7}{6}\Big)$
Length of perpendicular from (1, 1, 2) to the plane 2x - 2y + 4z + 5 = 0
$=\Bigg|\frac{2\times1-2\times1+4\times2+5}{\sqrt{(2)^2+(-2)^2+(4)^2}}\Bigg|\ \begin{pmatrix} \text{Length of perpendicular from P}(\text{x}_1,\text{y}_1,\text{z}_1)\text{ to the plane}\\ \text{ax}+\text{by}+\text{cz}+\text{d}=0=\Bigg|\frac{\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}\bigg| \end{pmatrix}$
$=\Big|\frac{2-2+8+5}{\sqrt{24}}\Big|$
$=\frac{13}{\sqrt{24}}$
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Question 764 Marks
the cartesian equation of a line are $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}.$ Find a vector equation for the line.
Answer
The cartesian equation of the given line is $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}.$
It can be re-written as
$\frac{\text{x}-5}{3}=\frac{\text{y}-(-4)}{7}=\frac{\text{z}-6}{2}.$
Thus, the given line passes through the point having position vector $\vec{\text{a}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}.$
We know that the vector equation of a line passing through a point with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}.$
Vector equation of the required line is
$\vec{\text{r}}=\big(5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}\big)$
Here, $\lambda$ is a parameter.
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Question 774 Marks
If the lines $\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{-2\text{k}}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{\text{k}}=\frac{\text{y}-2}{1}=\frac{\text{z}-3}{5}$ are perpendicular, find the value of k and, hence find the equation of the plane containing these lines.
Answer
We know that the lines $\frac{\text{x}-\text{x}_1}{\text{l}_1}=\frac{\text{y}-\text{y}_1}{\text{m}_1}=\frac{\text{z}-\text{z}_1}{\text{n}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{l}_2}=\frac{\text{y}-\text{y}_2}{\text{m}_2}=\frac{\text{z}-\text{z}_2}{\text{n}_2}$ are perpendicular if
l1l2 + m1m2 + n1n2 = 0
Here,
l1 = -3, m1 = -2k, n1 = 2, l2 = k, m2 = 1, n2 = 5
It is given that given are perpendicular.
⇒ l1l2 + m1m2 + n1n2 = 0
⇒ (-3)(k) + (-2k)(1) + (2)(5) = 0
⇒ -3k - 2k + 10 = 0
⇒ -5k = -10
⇒ k = 2
Substituting this value in the given equation of the lines, we get
$\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{2}\ ...(\text{i})$
$\frac{\text{x}-1}{2}=\frac{\text{y}-2}{1}=\frac{\text{z}-3}{5}\ ...(\text{ii})$
Finding the equation of the plane
Let the direction ratios of the required plane be proporional to a, b, c.
We know from (i) and (ii) that lines (i) and (ii) pass through the point (1, 2, 3) and the direction ratios of (i) and (ii) are proportional to -3, -4, 2 and 2, 1, 5 respectively.
Since the plane contains the lines (i) and (ii), the plane must pass through the point (1, 2, 3) and it must be parallel to the line.
So, the equation of the plane is
a(x - 1) + b(y - 2) + c(z - 3) = 0 ....(iii)
-3a - 4b + 2c = 0 ....(iv)
2a + b + 5c = 0 ....(v)
Solving (i), (ii) and (iii) we get
$\begin{vmatrix}\text{x}-1&\text{y}-2&\text{z}-3\\-3&-4&2\\2&1&5 \end{vmatrix}=0$
⇒ -22(x - 1) + 19(y - 2) + 5(z - 3) = 0
⇒ -22x + 19y + 5z = 31
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Question 784 Marks
Prove that the line $\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}\big)$ and $\vec{\text{r}}=\big(4\hat{\text{i}}-\hat{\text{k}}\big)+\mu\big(2\hat{\text{i}}+3\hat{\text{k}}\big)$ intersect and find their point of intersection.
Answer
The position vectors of two arbitrary points on the given lines are
$\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}\big)=(1+3\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}-\hat{\text{k}}$
$\big(4\hat{\text{i}}-\hat{\text{k}}\big)+\mu\big(2\hat{\text{i}}+3\hat{\text{k}}\big)=(4+2\mu)\hat{\text{i}}+0\hat{\text{j}}+(3\mu-1)\hat{\text{k}}$
If the lines intersect, then they have a common point. so, for some values of $\lambda$ and $\mu,$ we must have
$(1+3\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}-\hat{\text{k}}=(4+2\mu)\hat{\text{i}}+0\hat{\text{j}}+(3\mu-1)\hat{\text{k}}$
Equating the coefficients of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$1+3\lambda=4+2\mu\dots(1)$
$1-\lambda=0\dots(2)$
$3\mu-1=-1\dots(3)$
Solving (2) and (3), we get
$\lambda=1$
$\mu=0$
Substituting the valuse $\lambda=1$ and $\mu=0$ in (1), we get
$\text{LHS}=1+3\lambda$
$=1+3(1)$
$=4$
$\text{RHS}=4+2\mu$
$=4+2(0)$
$=4$
$\Rightarrow\text{LHS}=\text{RHS}$
Since $\lambda=1$ and $\mu=0$ satisfy (3), the given lines intersect.
Substituting $\mu=0$ in the second line, we get $\vec{\text{r}}=4\hat{\text{i}}+0\hat{\text{j}}-\hat{\text{k}}$ as the position vector of the point of intersection.
Thus, the coordinates of the point of intersection are (4, 0, -1).
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Question 794 Marks
Find the shortest distance between the lines
$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$ $\text{and}\ \vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big).$
Answer
$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$
$\vec{\text{a}}_1=4\hat{\text{i}}-\hat{\text{j}}\ \ \ \vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big)$
$\vec{\text{a}}_2=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}}_2=2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\cdot\big(\vec{\text{b}}_2\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1-\vec{\text{b}}_2\big|}\end{vmatrix}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}} &\hat{\text{j}}&\hat{\text{k}}\\1 & 2&-3\\2&4&-5 \end{vmatrix}$
$=\hat{\text{i}}(-10+12)-\hat{\text{j}}(-5+6)+\hat{\text{k}}(4-4)$
$=2\hat{\text{i}}-\hat{\text{j}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(-3\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}}\big)\cdot\big(2\hat{\text{i}}-\hat{\text{j}}\big)}{\sqrt{4+1}}\end{vmatrix}$
$=\begin{vmatrix}\frac{-6}{\sqrt{5}}\end{vmatrix}=\frac{6}{\sqrt{5}}$
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Question 804 Marks
Find the distance between the planes $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})+7=0$ and $\vec{\text{r}}\cdot(2\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}})+7=0$
Answer
The given plane are,
$\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=-7$
$\Rightarrow\text{x}+2\text{y}+3\text{z}=-7$
Multiplying this equation of the plane by 2, we get
$2​​\text{x}+4​\text{y}+6​\text{z}=-14\ ...(\text{i})$
and
$\vec{\text{r}}\cdot(2\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}})=-7$
$2​​\text{x}+4​\text{y}+6​\text{z}=-7\ ...(\text{ii})$
We know that distance between two planes ax + by + cz = d1 and ax + by + cz = d2 is $\frac{\big|\text{d}_2-\text{d}_1\big|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|-7-(-14)|}{\sqrt{2^2+4^2+6^2}}$
$=\frac{|7|}{\sqrt{4+16+36}}$
$=\frac{7}{\sqrt{56}}\text{ units}$
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Question 814 Marks
Find the distance of the point (2, 12, 5) from the point of intersection of the line $\vec{\text{r}}=2\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}}+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}})$ and $\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}})=0.$
Answer
The equation of the given line is $\vec{\text{r}}=2\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}}+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}})$
The position vector of any point on the line is
$\vec{\text{r}}=(2+3\lambda)​​\hat{\text{i}}+(-4+4\lambda)\hat{\text{j}}+(2-2\lambda)\hat{\text{k}}$
If this lies on the plane $\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}})=0,$ then
$\Big[(2+3\lambda)​​\hat{\text{i}}+(-4+4\lambda)\hat{\text{j}}+(2-2\lambda)\hat{\text{k}}\Big]\cdot(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}})=0$
$\Rightarrow(2+3\lambda)-2(-4+4\lambda)+(2+2\lambda)=0$
$\Rightarrow2+3\lambda+8-8\lambda+2+2\lambda=0$
$\Rightarrow3\lambda=12$
$\Rightarrow\lambda=4$
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Question 824 Marks
Find the distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along the line x = y = z.
Answer
The equation of line parallel to the line x = y = z and passing through the point (1, -5, 9) is
$\frac{\text{x}-1}{1}=\frac{\text{y}+5}{1}=\frac{\text{z}-9}{1}\ ...(\text{i})$
Any point on this line is of the form (k + 1, k - 5, k + 9)
If (k + 1, k - 5, k + 9) be the point of intersection of line (i) and the given plane, then
(k + 1) - (k - 5) + (k + 9) = 5
⇒ k = -10
So, the point of intersection of line (i) and the given plane is (-10 + 1, -10 - 5, -10 + 9) i.e., (-9, -15, -1).
$\therefore$ Required distance = Distance between (1, -5. 9) and (-9, -15, -1)
$=\sqrt{(1+9)^2+(-5+15)^2+(9+1)^2}$
$=\sqrt{3\times10^2}$
$=10\sqrt{3}\text{ units}$
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Question 834 Marks
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}})+9=0.$
Answer
Let a, b, c be the direction ratios of the given line.
Since the line passes through the point (1, 2, 3) is,
$\frac{\text{x}-1}{\text{a}}=\frac{\text{y}-2}{\text{b}}=\frac{\text{z}-3}{\text{c}}\ ...(\text{i})$
Since this line is perpendicular to the planer $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}})+9=0$ or x + 2y - 5z + 9 = 0, this line is parallel to the normal of the plane.
So, the direction ratios of the line are proportional to the direction ratios of the given plane.
So, $\frac{\text{a}}{1}=\frac{\text{b}}{2}=\frac{\text{c}}{-5}=\lambda$
$\text{a}=\lambda,\text{ b}=2\lambda,\text{ c}=-5\lambda$
Substituting these value in (i) we get
$\frac{\text{x}-1}{1}=\frac{\text{y}+1}{2}=\frac{\text{z}-2}{-5},$ which is the cartesian form of the line.
vector from
The given line passes through a point whose position vector is $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}.$ So, its equation in vector form is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\vec{\text{r}}=(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}})$
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Question 844 Marks
By computing the shortest distance determine whether the following pairs of lines intersect or not:
$\vec{\text{r}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
Answer
Given equations of lines are,
$\vec{\text{r}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}_1=\big(\hat{\text{i}}-\hat{\text{j}}\big),\vec{\text{b}}_1=\big(2\hat{\text{i}}+\hat{\text{k}}\big)$
and, $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{a}}_2=\big(2\hat{\text{i}}-\hat{\text{j}}\big),\vec{\text{b}}_2=\big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
We know that, shortest distance between lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\lambda\vec{\text{b}}_2$ is given by
$\text{S.D.}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|\dots(1)$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)=\big(2\hat{\text{i}}-\hat{\text{j}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}\big)$
$=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{i}}+\hat{\text{j}}$
$=\hat{\text{i}}$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&1\\1&1&-1 \end{vmatrix}$
$=\hat{\text{i}}(0-1)-\hat{\text{j}}(-2-1)+\hat{\text{k}}(2-0)$
$=-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=(\hat{\text{i}})\big(-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)$
$=(1)(-1)+(0)(3)+(0)(2)$
$=-1+0+0$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=-1$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-1)^2+(3)^2+(2)^2}$
$=\sqrt{1+9+4}$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{14}$
So, shortest distance between the given lines using equation (1) is,
$\text{S.D.}=\Big|\frac{-1}{\sqrt{14}}\Big|$
$=\frac{1}{\sqrt{14}}\text{ units}$
$\text{S.D.}\neq0$
Since, shortest distance between lines is not zero, so lines are not intersecting.
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Question 854 Marks
Find the distance of the point with position vector $-\hat{\text{i}}-5\hat{\text{j}}-10\hat{\text{k}}$ from the point of intersection of the line $\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+12\hat{\text{k}})$ with the plane $\vec{\text{r}}.(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5.$
Answer
The given equation of the line is,
$\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+12\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}=(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+2\lambda)\hat{\text{k}}$
The coordinated of any point on line are of the form $(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+2\lambda)\hat{\text{k}}$ or $(2+3\lambda,-1+4\lambda,2+2\lambda)$
Since this point lies on the plane $\vec{\text{r}}.(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5$
$\big[(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+2\lambda)\hat{\text{k}}\big].(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5$
$\Rightarrow 2+3\lambda+1-4\lambda+2+2\lambda-5=0$
$\Rightarrow \lambda=0$
So, the coordinates ofthe point are
$(2+3\lambda,-4+4\lambda,2+2\lambda)$
$=(2+1,-1+0,2+0)$
$=(2,-1,2)$
The coordinated of the point corresponding to the position vector $-\hat{\text{i}}-5\hat{\text{j}}-10\hat{\text{k}}$ are (-1, -5, -10).
Distance between (2, -1, 2) and (-1, -5, -10)
$=\sqrt{(-1-2)^2+(-5+1)^2+(-10-2)^2}$
$=\sqrt{9+16+144}$
$=13\text{ units}$
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Question 864 Marks
Show that the lines $\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})+\lambda(3\hat{\text{i}}-\hat{\text{j}})\ \text{and}\ \vec{\text{r}}=({4\hat{\text{i}}-\hat{\text{k}})+\mu(2\hat{\text{i}}+3\hat{\text{k}}})$intersect. Also find their point of intersection.
Answer
General points on the lines are
$(1+3\lambda)\hat{\text{i}}+(1-\lambda)\hat{\text{j}}-\hat{\text{k}}\ \ \&\ \ ({4+2\mu)\hat{\text{i}}+(3\mu-1)\hat{\text{k}}}$
lines intersect if
$1+3\lambda=4+2\mu\ \ \ \ \dots(1);$ $1-\lambda=0\ \ \ \ \dots(2);$   $3\mu-1=-1\ \ \ \ \dots(3)$ $\text{ for some }\lambda\ \&\ \mu$
From (2) & (3) λ =1, μ = 0
substituting in equation (1)
Since, 1 + 3(1) = 4 + 2 (0) is true $\therefore$ lines interset
Point of intersection is : $4\hat{\text{i}}-\hat{\text{k}}$ or (4, 0, -1)
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Question 874 Marks
Show that the lines $\frac{5-\text{x}}{-4}=\frac{\text{y}-7}{4}=\frac{\text{z}+3}{-5}$ and $\frac{\text{x}-8}{7}=\frac{2\text{y}-8}{2}=\frac{\text{z}-5}{3}$ are coplanar.
Answer
$\frac{5-\text{x}}{-4}=\frac{\text{y}-7}{4}=\frac{\text{z}+3}{-5}$
$\frac{\text{x}-5}{4}=\frac{\text{y}-7}{4}=\frac{\text{z}+3}{-5}\ ...(\text{i})$
$\frac{\text{x}-8}{7}=\frac{2\text{y}-8}{2}=\frac{\text{z}-5}{3}$
$\frac{\text{x}-8}{7}=\frac{\text{y}-4}{1}=\frac{\text{z}-5}{3}\ ....(\text{ii})$
Here, a1 = 4, b1 = 4, c1 = -5
a2 = 7, b2 = 1, c2 = 3
x1 = 5, y1 = 7, z1 = -3
x2 = 8, y2 = 4, z2 = 5
Condition for two lines to be coplanar,
$\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}=0$
$\therefore\begin{vmatrix}8-5&4-7&5+3\\4&4&-5\\7&1&3\end{vmatrix}$
$=\begin{vmatrix}3&-3&8\\4&4&-5\\7&1&3\end{vmatrix}$
$=3(12+5)+3(12+35)+8(4-28)$
$=3\times17+3\times47+8\times(-24)$
$=51+141-192$
$=192-192$
$=0$
$\therefore$ The lines are coplanar to each other.
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Question 884 Marks
Show that the lines $\frac{\text{x}}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}+3}{3}$ and $\frac{\text{x}-2}{2}=\frac{\text{y}-6}{3}=\frac{\text{z}-3}{4}$ intersect and find their point of intersection.
Answer
The coordinates of any point on the first line are given by
$\frac{\text{x}}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}+3}{3}=\lambda$
$\Rightarrow\text{x}=\lambda$
$\text{y}=2\lambda+2$
$\text{z}=3\lambda-3$
The coordinales of a general point on the first line are $\big(\lambda,2\lambda+2,3\lambda-3\big)$
Also, the coordinates of any point on the second line are given by
$\frac{\text{x}-2}{2}=\frac{\text{y}-6}{3}=\frac{\text{z}-3}{4}=\mu$
$\Rightarrow\text{x}=2\mu+2$
$\text{y}=3\mu+6$
$\text{z}=4\mu+3$
The coordinates of a general point on the second line are $\big(2\mu+2,3\mu+6,4\mu+3\big)$
It the lines intersect, then they have a common point. so, for some veluse of $\lambda$ and $\mu,$ we must have
$\lambda=2\mu+2,2\lambda+2=3\mu+6,3\lambda-3=4\mu+3$
$\Rightarrow\lambda-2\mu=2\dots(1)$
$2\lambda-3\mu=4\dots(2)$
$3\lambda-4\mu=6\dots(3)$
Solving (1) and (2), we get
$\lambda=2$ and $\mu=0$
Substituting $\lambda=2$ and $\mu=0$ in (3), we get
$\text{LHS}=3\lambda-4\mu$
$=3(2)-4(0)$
$=6$
$=\text{RHS}$
Since $\lambda=2$ and $\mu=0$ satisty the thied equation, the given lines intersect at (2, 6, 3).
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Question 894 Marks
Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinate of the point.
Answer
Equation of curve is y = x3 $\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{3x}^{2}$......................(i) 

= y – coordinate of the point $\Rightarrow$ 3x2 = y = x3 $\Rightarrow$ x2 (x – 3) = 0

x = 0, x = 3

When x = 0, y = 0, when x = 3, y = 27 

The points are (0,0), (3, 27).

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Question 904 Marks
Find the vector equations of the following planes in scalar product form $(\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}):$
$\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{k}})+\lambda\hat{\text{i}}+\mu(\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
Answer
Here, $\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{k}})+\lambda\hat{\text{i}}+\mu(\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
We know that, $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ represent a plane passing throught a point having position vector $\vec{\text{a}}$ and parallel to vectors $\vec{\text{b}}$ and $\vec{\text{c}}$
Here, $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}},\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}$
The given plane is perpendicular to a vector
$\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
 $=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&0&0\\1&-2&-1\end{vmatrix}$
$=\hat{\text{i}}(0-0)-\hat{\text{j}}(-1-0)+\hat{\text{k}}(-2-0)$
$=0\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{n}}=\hat{\text{j}}-2\hat{\text{k}}$
We know that vector equation of plane in scalar product form is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}\ ...(\text{i})$
Put $\vec{\text{n}}$ and $\vec{\text{a}}$ in equation (i),
$\vec{\text{r}}\cdot(\hat{\text{j}}-2\hat{\text{k}})=(2\hat{\text{i}}-\hat{\text{k}})(\hat{\text{j}}-2\hat{\text{k}})$
$\vec{\text{r}}\cdot(\hat{\text{j}}-2\hat{\text{k}})=(2)(0)+(0)+(1)+(-1)(-2)$
$\vec{\text{r}}\cdot(\hat{\text{j}}-2\hat{\text{k}})=0+0+2$
$\vec{\text{r}}\cdot(\hat{\text{j}}-2\hat{\text{k}})=2$
The equation in required form is,
$\vec{\text{r}}\cdot(\hat{\text{j}}-2\hat{\text{k}})=2$
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Question 914 Marks
Find the distance between the point (-1, -5, -10) and the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5.$
Answer
Any point on the line $\frac{\text{x} - 2}{3} = \frac{\text{y} + 1}{4} = \frac{\text{z} - 2}{12} = \text{is} (3\lambda + 2, 4\lambda - 1, 12\lambda + 2)$
If this is the point of intersection with plane $\text{x - y + z = 5}$
$\text{then 3} \lambda + 2 - 4\lambda + 1 + 12\lambda + 2 - 5 = 0 \Rightarrow \lambda = 0$
$\therefore$ Point of intersection is (2, -1, 2)
Required distance = $\sqrt{(2 + 1)^{2} +(-1 + 5)^{2} + (2 + 10)^{2} } = 13$
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Question 924 Marks
Find the vector equation of the following planes in non-parametric form.
$\vec{\text{r}}=(\lambda-2\mu)\hat{\text{i}}+(3-\mu)\hat{\text{j}}+(2\lambda+\mu)\hat{\text{k}}$
Answer
The given equation of the plane is,
$\vec{\text{r}}=(\lambda-2\mu)\hat{\text{i}}+(3-\mu)\hat{\text{j}}+(2\lambda+\mu)\hat{\text{k}}$
$\Rightarrow\vec{\text{r}}=(0\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}})+\lambda(\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}})+\mu(-2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})$
We know that equation $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}} $ represents a plane passing through a point whose position vector is $\vec{\text{a}}$ and parallel to t.
Here, $\vec{\text{a}}=0\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}},\hat{\text{c}}=-2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&0&2\\-2&-1&1\end{vmatrix}$
$=2\hat{\text{i}}-5\hat{\text{j}}-\hat{\text{k}}$
The vector equation of the plane in scalar product form is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}-5\hat{\text{j}}-\hat{\text{k}})=(0\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}})(2\hat{\text{i}}-5\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}-5\hat{\text{j}}-\hat{\text{k}})=0-15+0$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}-5\hat{\text{j}}-\hat{\text{k}})+15=0$
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Question 934 Marks
If $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ 2\hat{\text{i}}+5\hat{\text{J}},\ 3\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and $\hat{\text{i}}-6\hat{\text{j}}-\hat{\text{k}}$ respectively are the position vectors of points A, B, C and D, then find the angle between the straight lines AB and CD. Find whether $\vec{\text{AB}}$ and $\vec{\text{CD}}$ are collinear or not.
Answer
Given:
The position vector of A is $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}.$
The position vector of B is $2\hat{\text{i}}+5\hat{\text{j}}.$
Therefore, $\vec{\text{AB}}=(2-1)\hat{\text{i}}+(5-1)\hat{\text{j}}+(0-1)\hat{\text{k}}=\hat{\text{i}}+\hat4{\text{j}}-\hat{\text{k}}$
The position vector of C is $3\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and
The postion vector of D is $\hat{\text{i}}-6\hat{\text{j}}-\hat{\text{k}}.$
Therefore, $\vec{\text{CD}}=(1-3)\hat{\text{i}}+(-6-2)\hat{\text{j}}\\+(-1+3)\hat{\text{k}}=-2\hat{\text{i}}-\hat8{\text{j}}+2\hat{\text{k}}$
$\cos\theta=\frac{\vec{\text{AB}}.\vec{\text{CD}}}{|\vec{\text{AB}}||\vec{\text{CD}}|}$
$\Rightarrow\cos\theta=\frac{-2-32-2}{\sqrt{18}\sqrt{72}}=-1$
$\Rightarrow\theta=180^\circ$
Since, angle between Line AB and CD is 180°, therefore $\vec{\text{AB}}$ and $\vec{\text{CD}}$ are collinear.
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Question 944 Marks
Find the equation of the plane passing through (a, b, c) and parallel to the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2.$
Answer
Let the equation of a plane parallel to the given plane be
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{k}$
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{k}\ ...(\text{i})$
This passes through (a, b, c). So,
$(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{k}$
$\Rightarrow\text{k}=\text{a}+\text{b}+\text{c}$
Substituting this in (i), we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{a}+\text{b}+\text{c}$
$\text{x}+\text{y}+\text{z}=\text{a}+\text{b}+\text{c},$ Which is the equation of the required plane.
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Question 954 Marks
A plane passes through the point (1, -2, 5) and is perpendicular to the line joining the origin to the point $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}.$ Find the vector and cartesian forms of the equation of the plane.
Answer
The normal is passing throught the point A(0, 0, 0) and B(3, 1, -1). So,
$\vec{\text{n}}=\overrightarrow{\text{OP}}$
$=(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})-(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})$
$=3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
Since the plane passes throught (1, -2, 5)
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}}$
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 $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{n}}=4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},$ we get
$\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=(\hat{\text{i}}-2\hat{\text{j}}+5\hat{\text{k}})\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=3-2-5$
$\Rightarrow\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=-4$
Substituting $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=-4$
$\Rightarrow3\text{x}+\text{y}-\text{z}=-4$
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Question 964 Marks
Find the vector equations of the following planes in scalar product form $(\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}):$
$\vec{\text{r}}=(1+\text{s}-\text{t})\hat{\text{t}}+(2-\text{s})\hat{\text{j}}+(3-2\text{s}+2\text{t})\hat{\text{k}}$
Answer
Here, $\vec{\text{r}}=(1+\text{s}-\text{t})\hat{\text{t}}+(2-\text{s})\hat{\text{j}}+(3-2\text{s}+2\text{t})\hat{\text{k}}$
$\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})+\text{s}(\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})+\text{t}(-\hat{\text{i}}+2\hat{\text{k}})$
We know that, $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ represent a plane passing through a point having position vector $\vec{\text{a}}$ and parallel to vectors $\vec{\text{b}}$ and $\vec{\text{c}}.$
Here, $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}},\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{k}}$
The given plane is perpendicular to a vector
$\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
 $=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-1&-2\\-1&0&2\end{vmatrix}$
$=\hat{\text{i}}(-2-0)-\hat{\text{j}}(2-2)+\hat{\text{k}}(0-1)$
$\vec{\text{n}}=-2\hat{\text{i}}-\hat{\text{k}}$
We know that vector equation of plane in scalar product form is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}\ ...(\text{i})$
Put $\vec{\text{n}}$ and $\vec{\text{a}}$ in equation (i),
$\vec{\text{r}}\cdot(-2\hat{\text{i}}-\hat{\text{k}})=(-2\hat{\text{i}}-\hat{\text{k}})(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$
$\vec{\text{r}}\cdot(-2\hat{\text{i}}-\hat{\text{k}})=(-2)(1)+(0)(2)+(-1)(3)$
$\vec{\text{r}}\cdot(-2\hat{\text{i}}-\hat{\text{k}})=-2+0-3$
$\vec{\text{r}}\cdot(-2\hat{\text{i}}-\hat{\text{k}})=-5$
Multiplying both the sides by (-1),
$\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=5$
The equation in required form,
$\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=5$
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Question 974 Marks
Find the angle between the following pair of lines:
$\frac{\text{-x + 2}}{-2}=\frac{\text{y - 1}}{7}=\frac{\text{z + 3}}{-3}$ and $\frac{\text{x + 2}}{-1}=\frac{\text{2y - 8}}{4}=\frac{\text{z -5 }}{4}$
and check whether the lines are parallel or perpendicular.
Answer
$\frac{\text{x}-2}{2}=\frac{\text{y}-1}{7}=\frac{\text{z}+3}{-3}$
$\frac{\text{x}+2}{-1}=\frac{\text{y}-4}{2}=\frac{\text{z}-5}{4}$
The direction ratios of given lines are
2, 7, -3 and -1, 2, 4
Let $\theta$be the angle between these lines, then
$\cos\theta=\frac{2(-1)+7(2)+(-3)4}{\sqrt{4+49+9}\cdot\sqrt{1+4+16}}=0$
$\Rightarrow\theta=\frac{\pi}{2}$
Hence the lines are perpendicular to each other.
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Question 984 Marks
Find the shortest distance between the lines whose vector equations are:
$\vec{\text{r}}=\Big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\Big)+\lambda\Big(\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\Big)$
$\text{and}\ \vec{\text{r}}=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}+\mu\Big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\Big)$
Answer
Equation of the first line is $\vec{\text{r}}=\Big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\Big)+\lambda\Big(\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{r}}=\vec{\text{a}_1}+\lambda\vec{\text{b}_1},$
$\vec{\text{a}_1}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\ \text{and}\ \vec{\text{b}_1}=\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
Again equation of second line $\vec{\text{r}}=\Big(4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{r}}=\vec{\text{a}_2}+\mu\vec{\text{b}_2},$
$\vec{\text{a}_2}=4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\ \text{and}\ \vec{\text{b}_2}=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
Now shortest distance $(\text{d})=\frac{\Big|\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)\Big|}{\Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|}\ \ \ \ ...(\text{i})$
Here $\vec{\text{a}_2}-\vec{\text{a}_1}=\Big(4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\Big)-\Big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\Big)$
$=3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}_1}\times\vec{\text{b}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-3&2\\2&3&1\end{vmatrix}$
$=(-3-6)\hat{\text{i}}-(1-4)\hat{\text{j}}+(3+6)\hat{\text{k}}=-9\hat{\text{i}}+3\hat{\text{j}}+9\hat{\text{k}}$
$\Big|\vec{\text{b}}_1\times\vec{\text{b}}_2\Big|=\sqrt{(-9)^2+(3)^2+(9)^2}=\sqrt{171}=3\sqrt{19}$
$\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)=3\times(-9)+(3\times3)+(3\times9)$
$=-27+9+27=9$
Putting these values in eq.(i),
Shortest distance $(\text{d})=\frac{|9|}{3\sqrt{19}}=\frac{9}{3\sqrt{19}}=\frac{3}{\sqrt{19}}.$
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Question 994 Marks
Find the vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0.
Answer
The equation of the plane passing through the line of intersection of the given planes is,
$\text{x}+\text{y}+\text{z}-1+\lambda(2\text{x}+3\text{y}+4\text{z}-5)=0$
$(1+2\lambda)\text{x}+(1+3\lambda)\text{y}+(1+4\lambda)\text{z}-1-5\lambda=0\ ...(\text{i})$
This plane is perpendicular to x - y + z = 0. So,
$1+2\lambda-1(1+3\lambda)+1+4\lambda=0$ (Because a1a2 + b1b2 + c1c2 = 0)
$\Rightarrow1+2\lambda-1-3\lambda+1+4\lambda=0$
$\Rightarrow3\lambda+1=0$
$\Rightarrow\lambda=\frac{-1}{3}$
Substituting this in (i), we get
$\Big(1+2\Big(\frac{-1}{3}\Big)\Big)\text{x}+\Big(1+3\Big(\frac{-1}{3}\Big)\Big)\text{y}+\Big(1+4\Big(\frac{-1}{3}\Big)\text{z}-1-5\Big(\frac{-1}{3}\Big)\Big)=0$
$\Rightarrow\text{x}-\text{z}+2=0$
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Question 1004 Marks
$\text{Let } \vec{\text a} = \hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}, \vec{\text{b}} = \hat{\text{i}} \text{ and } \vec{\text{c}} = \text{c}_{1} \hat{\text{i}} + \text{c}_{2} \hat{\text{j}} + \text{c}_{3} \hat{\text{k}}, \text{then}$
  1. Let c1 = 1 and c2 = 2, find c3 which makes $\vec{\text{a}}, \vec{\text{b}} \text{ and }\vec{\text{c}} \text{ coplanar.}$
  2. If c2 = –1 and c3 = 1, show that no value of c1 can make $\vec{\text{a}}, \vec{\text{b}} \text{ and } \vec{\text{c}} \text{ coplanar}.$
Answer
$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ \text{c}_{1} & \text{c}_{2} & \text{c}_{3} \end{vmatrix} = \text{c}_{2} - \text{c}_{3}$

  1. $\text{c}_{1} = 1, \text{ c}_{2} = 2$

$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 2 - \text{c}_{3}$

$\therefore \vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ are coplanar} [\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 0 \Rightarrow \text{c}_{3} = 2$

  1. $\text{c}_{2} = -1, \text{c}_{3} = 1$

$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \text{c}_{2} - \text{c}_{3} = -2 \neq 0$

$\Rightarrow \text{No value of }\text{c}_{1} \text{can make }\vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ coplanar}.$ 

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