5 Marks Questions

Question 15 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 $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are $x_2 x_1, y_2 y_1, z_2 z_1. 2$
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 25 Marks

Find the angle between the lines whose direction cosines are given by the equations: $l + m +n = 0$ and $l^2 + m^2 + n^2 = 0$

Answer

Given that, $l + m + n = 0 l^2 + m^2 + n^2 = 0$
From equation $(1), l = -(m + n)$
Put the value of $l$ in equation $(2),$
$[-(m + n)]^2 + m^2 - n^2 = 0 (m + n)^2 + m^2 - n^2 = 0 m^2+ n^2 +$$ 2mn + m^2 - n^2 = 0 2m^2 + 2mn = 0 2m(m + n) = 0 m = 0, m + n = 0 m = -n$ and $m = 0$
Put the value of $m = -n$ in equation $(1) l = -(m +n) = -(0 + n) l = -n$
Thus, the direction ratios are proportional to $0, -n, n$ and $-n, 0, n$
$\Rightarrow 0, -1 1$ and $-1, 0, 1$
So, vectors parallel to these lines are $\vec{a}=0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{b}=-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}}$ respectively.
Let, $\theta$ be the angle between the $\vec{a}$ and $\vec{b}$
So, $\cos\theta=\frac{\vec{a}\times\vec{b}}{\big|\vec{a}\big|\big|\vec{b}\big|}$
$\vec{a}=0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{b}=-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}}$
respectively. $\cos\theta=\frac{(0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\times(-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}})}{\sqrt{0^2+(-1)^2+(1)^2}\sqrt{(-1)^2+(0)^2+(1)^2}}$
$=\frac{(0)(-1)+(-1)(0)+(1)(1)}{\sqrt{1+1}\sqrt{1+1}}$
$=\frac{0+0+1}{\sqrt{2}\times\sqrt{2}}$
$=\frac{1}{2}$
$\theta=\cos^{-1}\Big(\frac{1}{2}\Big)$
$\theta=\frac{\pi}{3}$
So, angle between the lines $=\frac{\pi}{3}$.

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Question 35 Marks

Using direction ratios show that the points A(2, 3, -4) B(1, - 2, 3) and C(3, 8, -11) are collinear.

Answer

Here A(2, 3, -4), B(1, -2, 3) and C(3, 8, -11).
Direction ratios of AB = (1 - 2, -2 - 3, 3 + 4) = (-1, -5, 7)
Direction ratios of BC = (3 - 1, 8 + 2, -11 - 3) = (2, 10, -14)
Here, the respective direction consines of AB and AC,
$\frac{-1}{2}=\frac{-5}{10}=\frac{7}{-14}$ are proportional.
Also, Bis the common point between the two lines,
$\therefore$ The points A(2, 3, -4) B(1, -2, 3) and C(3, 8, -11) are collinear.

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Question 45 Marks

Find the angle between the lines whose direction cosines are given by the equations: $2l - m + 2n = 0$ and $mn + nl + lm = 0$

Answer

Given that,
$2l - m + 2n = 0 .....(1)$
$mn + nl + lm = 0 .....(2)$
From equation $(1),$
$2l - m + 2n = 0$
$m = 2l + 2n$
Put the value of m in equation $(2),$
$mn + nl + lm = 0$
$(2l + 2n) n + nl + l(2l + 2n) = 0$
$2ln + 2n^2 +nl +2l^2 + 2ln = 0$
$2l^2 + 5ln + 2n^2 = 0$
$2l^2 + 4ln + ln + 2ln^2 = 0$
$2l (l + 2n) + n(l + 2n) = 0$
$(1 + 2n) (2l = n) = 0$
$l + 2n = 0 or 2l + n = 0$
$l = -2n$ or $\text{l}=-\frac{\text{n}}{2}$
Put the value of $l = -2n$ in equation $(1)$
$2l - m + 2n = 0$
$2 (-2n) - m + 2n = 0$
$-4n - m + 2n = 0$
$-2n - m = 0$
$-2n = m$
$m = -2n$
Again, put the value of $\text{l}=-\frac{1}{2}$ in equation $(1)$
$2l - m + 2n = 0$
$2\Big(-\frac{1}{2}\text{n}\Big)-\text{m}+2\text{n}=0$
$-n - m + 2n = 0$
$-m + n = 0$
$-m = -n$
$m = n$
So, direction cosines of the lines are given by,
$-2n, -2n, n$ or $-\frac{1}{2},\text{n},\text{n},\text{n}$
$-2, -2, 1$ or $-\frac{1}{2},1,1$
So, vectors parallel to these lines
$\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=-\frac{1}{2}\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ respectively.
Let, $\theta$ be the angle between the $\vec{\text{a}}$ and $\vec{\text{b}},$
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{\big(-2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)\times\Big(-\frac{1}{2}\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)}{\sqrt{(-2)^2+(-2)^2+(1)^2}\sqrt{\Big(-\frac{1}{2}^2\Big)+(1)^2+(1)^2}}$
$=\frac{(-2)\big(-\frac{1}{2}\big)+(-2)(1)+(1)(1)}{\sqrt{4+4+1}\sqrt{\frac{1}{4}1+1}}$
$=\frac{1-2+1}{\sqrt{9}\sqrt{\frac{9}{4}}}$
$=\frac{0}{3\times\frac{3}{2}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
$\theta=\frac{\pi}{2}$
So, angle between the lines $=\frac{\pi}{2}$.

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Question 55 Marks

Find the angle between the vectors whose direction cosines are proportional to 2, 3, -6 and 3, -4, 5.

Answer

Let $\vec{\text{a}}$ be a vector with direction ratios 2, 3, -6.
$\Rightarrow\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
Let $\vec{\text{b}}$ be a vector with direction ratios 3, -4, 5.
$\Rightarrow\vec{\text{b}}=3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$
Let $\theta$ be the angle between the given vectors.
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}).(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})}{\big|2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}\big|\big|3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big|}$
$=\frac{6-12-30}{\sqrt{4+9+36}\sqrt{9+16+25}}$
$=\frac{-36}{\sqrt{49}\sqrt{50}}$
$=\frac{-36}{35\sqrt{2}}$
Rationalising the result, we get
$\cos\theta=-\frac{18\sqrt{2}}{35}$
$\therefore\theta=\cos^{-1}\Big(-\frac{18\sqrt{2}}{35}\Big)$
Thus, the angle between the given vectors measures $\cos^{-1}\Big(-\frac{18\sqrt{2}}{35}\Big)$.

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Question 65 Marks

Find the angle between the lines whose direction cosines are given by the equations $l + 2m + 3n = 0$ and $3lm - 4ln + mn = 0$

Answer

Given that,
$l + 2m + 3n = 0 .....(1)$
$3lm - 4ln + mn = 0 .....(2)$
From equation $(1),$
$l + 2m + 3n = 0$
$l = -2m - 3n$
Put the value of l in equation $(2),$
$3lm - 4ln + mn = 0$
$3(-2m - 3n) m - 4(-2m - 3n) n + mn = 0$
$-6m^{2 }- 9nm + 8mn + 12n^2 + mn = 0$
$-6m^2 + 12n^2 = 0$
$m^2 = 2n^2$
$\text{m}=\pm\sqrt{2\text{n}^2}$
$\text{m}=\text{n}\sqrt{2}$ or $\text{m}=-\text{n}\sqrt{2}$
Put $\text{m}=\text{n}\sqrt{2}$ in equation $(1)$
$l + 2m + 3n = 0$
$\text{l}+2\big(\text{n}\sqrt{2}\big)+3\text{n}=0$
$\text{l}+\text{n}\big(2\sqrt{2}+3\big)=0$
$\text{l}+-\big(2\sqrt{2}+3\big)\text{n}$
Again, $\text{m}=-\sqrt{2\text{n}}$ in equation $(1)$
$l + 2m + 3n = 0$
$\text{l}+2\big(-\sqrt{2\text{n}}\big)+3\text{n}=0$
$\text{l}-2\sqrt{2\text{n}}+3\text{n}=0$
$\text{l}+\text{n}\big(-2\sqrt{2\text{n}}+3\big)=0$
$\text{l}=\big(2\sqrt{2\text{n}}-3\big)\text{n}$
Thus, direction cosines of the lines are given by,
$-\big(2\sqrt{2}+3\big)\text{n},\sqrt{2\text{n}},\text{n}$ or $\big(2\sqrt{2}-3\big)\text{n},\sqrt{2\text{n}},\text{n}$
$-\big(2\sqrt{2}+3\big)\text{n},\sqrt{2},1$ or $\big(2\sqrt{2}-3\big)\text{n},-\sqrt{2},1$
So, vectors parallel to these lines are
$\vec{\text{a}}=-\Big(2\sqrt{2}+3\Big)\hat{\text{i}}+\sqrt{2}\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\Big(2\sqrt{2}-3\Big)\hat{\text{i}}-\sqrt{2}\hat{\text{j}}+\hat{\text{k}}$ respectively.
Let, $\theta$ be the angle between the lines,
then,
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\big|\times\big|\vec{\text{b}}\big|}$
$=\frac{-\big(2\sqrt{2}+3\big)\times{\big(2\sqrt{2}-3\big)+\big(\sqrt{2}\big)}\times\big(-\sqrt{2}\big)+(1)(1)}{\sqrt{\big(2\sqrt{2}+3\big)^2+\big(-\sqrt{2}\big)^2+(1)^2}\sqrt{\big(2\sqrt{2}-3\big)^2+\big(-\sqrt{2}\big)^2+(1)^2}}$
$=\frac{-(8-9)-2+1}{\sqrt{8+9+12\sqrt{2}+2+1\sqrt{8+9-12\sqrt{2}+2+1}}}$
$=\frac{-(-1)-2+1}{\sqrt{20+12\sqrt{2}}\sqrt{20-12\sqrt{2}}}$
$=\frac{1-2+1}{\sqrt{20+12\sqrt{2}}\sqrt{20-12\sqrt{2}}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
$\theta=\frac{\pi}{2}$
Angle between the lines $=\frac{\pi}{2}$.

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Question 75 Marks

Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.

Answer

The vectors, represented by these are
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
and $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
Let, $\theta$ be the angle between the lines,
then,
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\Big|\vec{\text{a}}\Big|\Big|\vec{\text{b}}\Big|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{(2)^2+(3)^2+(6)^2}\sqrt{(1)^2+(2)^2+(2)^2}}$
$=\frac{(2)(1)+(3)(2)+(6)(2)}{\sqrt{4+9+36}\sqrt{1+4+4}}$
$=\frac{2+6+12}{\sqrt{49}\sqrt{9}}$
$=\frac{20}{7\times3}$
$\cos\theta=\frac{20}{21}$
$\theta=\cos^{-1}\Big(\frac{20}{21}\Big)$
Angle between the lines $=\cos^{-1}\Big(\frac{20}{21}\Big)$.

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Question 85 Marks

Find the angle between the vectors with direction ratios proportional to 1, -2, 1 and 4, 3, 2.

Answer

Let $\vec{\text{a}}$ be a vector with direction ratios 1, -2, 1.
$\Rightarrow\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
Let $\vec{\text{b}}$ be a vector with direction ratios 4, 3, 2.
$\Rightarrow\vec{\text{b}}=4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
Let $\theta$ be the angle between the given vectors.
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}).(4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\big|\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big|\big|4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\big|}$
$=\frac{4-6+2}{\sqrt{1+4+1}\sqrt{16+9+4}}$
$=\frac{0}{\sqrt{6}\sqrt{29}}$
$=0$
$\therefore\theta=\frac{\pi}{2}$
Thus, the angle between the given vectors measures $\frac{\pi}{2}$.

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Question 95 Marks

Find the angle between the lines whose direction cosines are given by the equations: $2l + 2m - n = 0, mn + ln + lm = 0$

Answer

The given equation are,
$2l + 2m - n = 0 .....(1)$
$mn + ln + lm = 0 .....(2)$
From $(1),$ We get $n = 2l + 2m$.
Putting $n = 2l + 2m$ in $(2),$ We get
$m(2l + 2m) + l(2l + 2m) + lm = 0$
$2lm + 2m^2 + 2l^2 + 2ml + lm = 0$
$2ml^2 + 5lm + 2l^2 = 0$
$2m^2 + 4lm + lm + 2l^2 = 0$
$(2m + l) (m + 2l) = 0$
$\Rightarrow\text{m}=-\frac{1}{2}$ or $\text{m}=-2\text{l}$
By putting $\text{m}=-\frac{\text{l}}{2}$ in $(1)$ we get $n = l$
By putting $m = 2l$ in $(i)$ we get $n = -2l$
So, vector parallel to these lines are
$\vec{\text{a}}=\hat{\text{i}}-\frac{1}{2}\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$ respectively.
If $\theta$ is the angle between the lines, then $\theta$ is also the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$
then,
$\cos\theta=\frac{\vec{\text{a}}\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{1+1-2}{\sqrt{1+\frac{1}{4}+1}\sqrt{1+4+9}}=0$
$\theta=\cos^{-1}(0)=\frac{\pi}{2}$.

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Question 105 Marks

Find the direction cosines of the lines, connected by the relations: $l + m + n = 0$ and $\frac{2}{\text{m}}+\frac{2}{\text{n}}-\text{mn}=0$.

Answer

Given:
$l + m + n = 0 ......(1)$
$2lm + 2ln + nm = 0 ......(2)$
From $(1),$ we get
$l = m - n$
Substituting $l = -m - n $ in $(2),$ we get
$2(-m - n) m + 2(-m - n)n - mn = 0$
$\Rightarrow -2m^2 - 2mn - 2mn - 2n^2 - mn = 0$
$\Rightarrow 2m^2 + 2n^2 + 5mn = 0$
$\Rightarrow (m + 2n) (2m + n) = 0$
$\Rightarrow\text{m}=-2\text{n},-\frac{\text{n}}{2}$
If $m = -2n,$ then from $(1),$ we get $l = n.$
If $\text{m}=-\frac{\text{n}}{2},$ then from (1), we get $\text{l}=-\frac{\text{n}}{2}.$
Thus, the direction ratios of the two lines are proportional to $n, - 2n, n$ and $-\frac{\text{n}}{2},-\frac{\text{n}}{2},\text{n},$
i.e.$1,-2, 1$ and $-\frac{1}{2},-\frac{1}{2},1$
Hence, their direction cosines are
$\pm\frac{1}{\sqrt{6}},\pm\frac{-2}{\sqrt{6}},\pm\frac{1}{\sqrt{6}}$
$\pm\frac{-1}{\sqrt{6}},\pm\frac{-1}{\sqrt{6}},\pm\frac{2}{\sqrt{6}}$.

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Question 115 Marks

Find the angle between the lines whose direction ratios are proportional to a, b, c and b - c, c - a, a - b.

Answer

Given, that the direction ratios of lines are proportional to a, b, c and b - c, c - a, a - b.
Let, $\vec{\text{x}}$ and $\vec{\text{y}}$ be the vector parallel to these lines respectively, so
$\vec{\text{x}}=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$
And, $\vec{\text{y}}=(\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}$
Let, $\theta$ be the angle between $\vec{\text{x}}$ and $\vec{\text{y}}$, so,
$\cos\theta=\frac{\vec{\text{x}}\times\vec{\text{y}}}{\big|\vec{\text{x}}\big|\big|\vec{\text{y}}\big|}$
$=\frac{(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})[(\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}]}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{{(\text{b}-\text{c})^2+(\text{c}-\text{a})^2+(\text{a}-\text{b})^2}}}$
$=\frac{(\text{a})(\text{b}-\text{c})+\text{b}(\text{c}-\text{a})+\text{c}(\text{a}-\text{b})}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2\sqrt{\text{b}^2+\text{c}^2-2\text{bc}+\text{c}^2+\text{a}^2-2\text{ac}+\text{a}^2+\text{b}^2-2\text{ab}}}}$
$\cos\theta=\frac{\text{ab}-\text{ac}+\text{bc}-\text{ba}+\text{ca}-\text{bc}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
$\theta=\frac{\pi}{2}$
Angle between the lines $=\frac{\pi}{2}$

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Question 125 Marks

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, -4), (-1, 1, 2) and (-5, -5, -2).

Answer

A(3, 5, -4), B(-1, 1, 2) and C(-5, -5, -2) The direction ratios of the side AB = (-1 - 3, 1 - 3, 2 + 4) = (-4, -4 ,6) Direction cosines of AB will be $\frac{-4}{\sqrt{(-4)^2+(-4)^2+6^2}},\frac{-4}{\sqrt{(-4)^2+(-4)^2+6^2}},\frac{6}{\sqrt{(-4)^2+(-4)^2+6^2}}$ $=\frac{-4}{\sqrt{68}},\frac{-4}{\sqrt{68}},\frac{6}{\sqrt{68}}$ $=\frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}$ The direction ratios of the side BC = (-5 + 1, -5 - 1, -2 - 2) = (-4, -6, -4) Direction cosines of BC will be $\frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}$ $=\frac{-4}{\sqrt{68}},\frac{-6}{\sqrt{68}},\frac{-4}{\sqrt{68}}$ $=\frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}$ The direction ratios of the side AC = (-5 - 3, -5 - 5, -2 + 4) = (-8, -10, 2)Direction cosines of AC will be
$\frac{-8}{\sqrt{(-8)^2+(-10)^2+2^2}},\frac{-10}{\sqrt{(-8)^2+(-10)^2+2^2}},\frac{2}{\sqrt{(-8)^2+(-10)^2+2^2}}$ $=\frac{-8}{\sqrt{168}},\frac{-10}{\sqrt{168}},\frac{2}{\sqrt{168}}$ $=\frac{-4}{\sqrt{42}},\frac{-5}{\sqrt{42}},\frac{1}{\sqrt{42}}$.

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Question 135 Marks

Show that the points $(2, 3, 4), (-1, -2, 1), (5, 8, 7)$ are collinear.

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 joining the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are $x_{2 }- x_{1, }y_{2 }- y_{1, }z_{2 }- z_{1.}$
The direction ratios of $AB$ are $(-1 - 2), (-2 - 3), (1 - 4),$
i.e. $-3, -5, -3.$
The direction ratios of $BC$ are $(5 - (-1)), (8 - (-2)), (7 - 1),$
i.e. $6, 10, 6.$
It can be seen that the direction ratios of $BC$ are $-2$ times that of $AB,$
i.e. they are proportional. Therefore, $AB$ is parallel to $BC.$
Since point $B$ is common in both $AB$ and $BC,$ points $A, B,$ and $C$ are collinear.

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