3 Marks Question

Question 13 Marks

Find the direction cosines of the line passing through two points $(-2, 4, -5)$ and $(1, 2, 3)$.

Answer

The direction consines of the line passing through two points $P x_1, y_1, z_1$, and $Q (x_2, y_2, z_2)$ are $\frac{\text{x}_2-\text{x}_1}{\text{PQ}},\frac{\text{y}_2-\text{y}_1}{\text{PQ}},\frac{\text{z}_2-\text{z}_1}{\text{PQ}}.$
Here,
$\text{PQ}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2+(\text{z}_2-\text{z}_1)^2}$
$\text{P}=2,4,-5$
$\text{Q}=1,2,3$
$\therefore\text{PQ}=1-(-2)^2+(2-4)^2+[3-(-5)]^2=\sqrt{77}$
Thus, the direction cosines of the line joining two points are
$\frac{1-(-2)}{\sqrt{77}},\frac{2-4}{\sqrt{77}},\frac{3-(-5)}{\sqrt{77}},\text{i.e.}\frac{3}{\sqrt{77}}77,\frac{-2}{\sqrt{77}}77,\frac{8}{\sqrt{77}}.$

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Question 23 Marks

Show that the line joining the origin to the point $(2, 1, 1)$ is perpendicular to the line determined by the points $(3, 5, -1)$ and $(4, 3, -1).$

Answer

Here,
$A(0, 0, 0)$ and $B(2, 1, 1)$
$C(3, 5, -1)$ and $D(4, 3, -1)$
Direction ratios of line $AB$
$a_1 = 2, b_1 = 1, c_1= 1$
Direction ratios of line $CD$
$a_2 = 2, b_2 = -2, c_2= 0$
Now,
$a_1a_2+ b_1b_2+ c_1c_2$
$= (2)(1) + (1)(-2) + (1)(0)$
$= 2 - 2 + 0$
$= 0$
Since, $a_1a_2+ b_1b_2+ c_1c_2= 0,$ lines are perpendicular.

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Question 33 Marks

Show that the line through points $(4, 7, 8)$ and $(2, 3, 4)$ is parallel to the line throught the points $(-1, -2, 1)$ and $(1, 2, 5).$

Answer

Suppose the points are $A(2, 3, 4), B(-1, -2, 1)$ and $C(5, 8, 7).$
We know that the direction ratios of the line passing through the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are $x_2- x_{1,}y_2- y_{1,}z_2- z_{1.}$
Let the first two points be $A(4, 7, 8)$ and $B(2, 3, 4).$
Thus, the direction ratios of AB are $(2 - 4), (3 - 7), (4 - 8),$ i.e. $-2, -4, -4.$
Similarly, Let the other two points be $C (-1, -2, 1)$ and $D (1, 2, 5).$
Thus, the direction ratios of $CD$ are $[1 - (-1)], [2 - (-2)], (5 - 1),$ i.e. $2, 4, 4.$
It can be seen that the direction ratios of $CD$ are $-1$ times that of $AB,$ i.e. they are proportional.
Therefore, $AB$ and $CD$ are parallel lines.

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Question 43 Marks

A line makes an angle of 60° with each of X-axis and Y-axis. Find the acute angle made by the line with Z-axis.

Answer

It is given that a line makes an angle of 60° with both x-axis and y-axis.
Suppose the line makes an angle of $\alpha$ with the z-axis.
$\Rightarrow\text{l}=\cos60^\circ=\frac{1}{2}\text{m}$
$=\cos60^\circ=\frac{1}{2}\text{n}=\cos\alpha$
We know $\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\Big(\frac{1}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2+(\cos\alpha)^2=1$
$\Rightarrow\frac{1}{4}+\frac{1}{4}+\cos ^2\alpha=1$
$\Rightarrow\cos\alpha=\frac{1}{\sqrt{2}}$
$\Rightarrow\alpha=45^\circ$
Thus, the line makes an angle of 45° with the z-axis.

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Question 53 Marks

If a unit vector $\vec{\text{a}}$ makes an angle $\frac{\pi}{3}$ with $\hat{\text{i}},\frac{\pi}{4}$ with $\hat{\text{j}}$ and an acute angle $\theta$ with $\hat{\text{k}}$, and ,then find the value of $\theta$.

Answer

Scince a unit vector makes an angle of $\frac{\pi}{3}$ with$\hat{\text{i}}$, $\frac{\pi}{4}$ with $\hat{\text{j}}$ andan acute angle $\theta$ with $\hat{\text{k}},\text{l}=\cos\frac{\pi}{3}$ or $\frac{\pi}{4}$ or $\frac{1}{\sqrt{2}}$and $\text{n}=\cos\theta$.
We know
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\Big(\frac{1}{2}\Big)^2+\Big(\frac{1}{\sqrt{2}}\Big)^2+\cos^2\theta=1$
$\Rightarrow\frac{1}{4}+\frac{1}{2}+\cos^2\theta$
$\Rightarrow\cos^2\theta=\frac{1}{4}$
$\Rightarrow\cos^2\theta=\frac{1}{2}$
$\Rightarrow\frac{\pi}{3}$
Thus, the vector $\vec{\text{a}}$ makes an angle of $\frac{\pi}{3}$ with $\hat{\text{k}}$.

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Question 63 Marks

Show that the line through points $(1, -1, 2)$ and $(3, 4, -2)$ is perpendicular to the line throught the points $(0, 3, 2)$ and $(3, 5, 6).$

Answer

We know that two lines with direction ratios $a_1,b_1, c_2$ and $a_2, b_2, c_2$ are pependicular if $a_1a_2 + b_1b_2 + c_1c_2 = 0.$
The direction ratios of the line passing through the points $(1, -1, 2)$ and $(3, 4, -2)$ are $(3 - 1), [4 - (-1)], (-2 - 2),$
i.e. $\Rightarrow a_1= 2, b_1= 2, c_1= -4$
Similarly, the direction ratios of the line passing through the points $(0, 3, 2)$ and $(3, 5, 6)$ and $(3 - 0), (5 - 3), (6 - 2),$
i.e. $\Rightarrow a_2= 3, b_2= 2, c_2= 4$
$\therefore a_1a_2+ b_1b_2+ c_1c_2= 2 \times 3 + 5 \times 2 (-4) \times 4 = 6 + 10 - 16 = 0$
Thus the line through the points $(1, -1, 2)$ and $(3, 4, -2)$ is perpendicular to the line throught the points $(0, 3, 2)$ and $(3, 5, 6).$

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Question 73 Marks

Write the angle between the lines whose direction ratios are perportional to 1, -2, 1 and 4, 3, 2.

Answer

The direction ratios of the first line are 1, -2, 1 and the direction ratios of the second line are 4, 3, 2.
Let $\theta$ be the angle between these two lines.
Now,
$\cos\theta =\Bigg|\frac{1(4)+(-2)(3)+1(2)}{\sqrt{(1)^2+(-2)^2+(1)^2}\sqrt{(4)^2+(3)^2+(2)^2}}\Bigg|$
$=\Bigg|\frac{4+6+2}{\sqrt{1+4+1}\sqrt{16+9+4}}\Bigg|$
$=\frac{0}{\sqrt{6}\sqrt{29}}$
$=0$
$\Rightarrow\theta=\frac{\pi}{2}$
Hence, the required angle is$\frac{\pi}{2}$.

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