MCQ

MCQ 11 Mark

The projections of a line segment on $x, y$ and $z$ axes are $12, 4$ and $3$ respectively. The length and direction cosines of the line segment are:

✓
$13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$
B
$19;\frac{12}{19},\frac{4}{19},\frac{3}{19}$
C
$11;\frac{12}{11},\frac{14}{11},\frac{3}{11}$
D
$\text{None of these}$

Answer

Correct option: A.

$13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$

If a line makes angles $\alpha,\beta$ and $\gamma$ with the axes, then $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\dots(1)$
Let $r$ be the length of the line segment. then,
$\text{r}\cos\alpha=12,\text{r}\cos\beta=4,+\cos\gamma=3\dots(2)$
$\Rightarrow\big(\text{r}\cos\alpha\big)^2+\big(\text{r}\cos\beta\big)^2+\big(\text{r}\cos\gamma\big)^2=12^2+4^3+3^2$
$\Rightarrow\text{r}^2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)=169$
$\Rightarrow\text{r}^2(1)=169 [$From $(1)]$
$\Rightarrow\text{r}=\sqrt{169}$
$\Rightarrow\text{r}=\pm13$
$\Rightarrow\text{r}=13 ($Since length cannot be negative$)$
$($Since legth cannot be negative$)$
Substituting $r = 13$ in $(2),$ we get
$\cos\alpha=\frac{12}{13},\cos\beta\frac{4}{13},\cos\gamma=\frac{1}{13}$
Thus, the direction cosines of the line are $\frac{12}{13},\frac{4}{13},\frac{1}{13}$

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MCQ 21 Mark

The angle between the straight lines $\frac{\text{x}+1}{2}=\frac{\text{y}-2}{5}=\frac{\text{z}+3}{4}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{-3}$ is:

A
$45^\circ$
B
$30^\circ$
C
$60^\circ$
✓
$90^\circ$

Answer

Correct option: D.

$90^\circ$

We have
$\frac{\text{x}+1}{2}=\frac{\text{y}-2}{5}=\frac{\text{z}+3}{4}$
$\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{-3}$
The direction of the given lines are propotional to $2, 5, 4$ are $1, 2, -3.$
The given lines are parallel to the vectors $\vec{\text{b}}_1=2\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}.$
Let $\theta$ be the angle between the given lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(2\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)}{\sqrt{2^2+5^2+4^2}\sqrt{1^2+2^2+(-3)^2}}$
$=\frac{2+10-12}{\sqrt{45}\sqrt{14}}$
$\Rightarrow\theta=90^\circ$

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MCQ 31 Mark

The direction ratios of the line $x - y + z - 5 = 0 = x - 3y - 6$ are proportional to:

✓
$3,1,-2$
B
$2,-4,1$
C
$\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}}$
D
$\frac{2}{\sqrt{41}},\frac{-4}{\sqrt{41}},\frac{1}{\sqrt{41}}$

Answer

Correct option: A.

$3,1,-2$

We have
$x - y + z - 5 = 0 = x - 3y - 6$
$\Rightarrow x - 3y - 6=0$
$x - y + z - 5 = 0$
$\Rightarrow x = 3y + 6 .....(1)$
$x - y + z - 5 = 0.....(2)$
From $(1)$ and $(2),$ we get
$3y + 6 - y + z - 5 = 0$
$\Rightarrow 2y + z + 1 = 0$
$\Rightarrow\text{y}=\frac{-\text{z}-1}{2}$
$\text{y}=\frac{\text{x}-6}{3}$ [From (1)]
$\therefore\frac{\text{x}-6}{3}=\text{y}=\frac{-\text{z}-1}{2}$
So, the given equation can be re$-$witten as
$\frac{\text{x}-6}{3}=\frac{\text{y}}{1}=\frac{\text{z}+1}{-2}$
Hence, the direction ratios the given line are proportional to $3, 1, -2.$

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MCQ 41 Mark

If a line makes angle $\frac{\pi}{3}$ and $\frac{\pi}{4}$ with $x-$axis and $y-$axis respectively, then the angle made by the line with $z-$axis is:

A
$\frac{\pi}{2}$
✓
$\frac{\pi}{3}$
C
$\frac{\pi}{4}$
D
$\frac{5\pi}{12}$

Answer

Correct option: B.

$\frac{\pi}{3}$

If a line makes angles $\alpha,\beta$ and $\gamma$ with the axcs, then $\cos2\alpha+\cos2\beta+\cos2\gamma=1.$
Here,
$\alpha=\frac{\pi}{3}$
$\beta=\frac{\pi}{4}$
Now,
$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$
$\Rightarrow\cos^2\frac{\pi}{3}+\cos^2\frac{\pi}{4}+\cos^2\gamma=1$
$\Rightarrow\frac{1}{4}+\frac{1}{2}+\cos^2\gamma=1 $
$\Rightarrow\cos^2\gamma=1-\frac{3}{4}$
$\Rightarrow\cos^2\gamma=\frac{1}{4}$
$\Rightarrow\cos\gamma=\frac{1}{2}$
$\Rightarrow\gamma=\frac{\pi}{3}$

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MCQ 51 Mark

The shortest distance between the lines $\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}$ and, $\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}$ is:

A
$\sqrt{30}$
B
$2\sqrt{30}$
C
$5\sqrt{30}$
✓
$3\sqrt{30}$

Answer

Correct option: D.

$3\sqrt{30}$

We have
$\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}\dots(1)$
$\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}\dots(2)$
We know that line $(1)$ passes through the point $(3, 8, 3)$ and has direction ratios proportional to $3, -1, 1.$
Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1,$
where $\vec{\text{a}}_1=3\hat{\text{i}}+8\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{b}}_1=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}.$
Also, line $(2)$ passes through the point $(3, -7, 6)$ and has direction ratios proprtional to $-3, 2, 4.$
Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2,$ where $\vec{\text{a}}_2=-3\hat{\text{i}}-7\hat{\text{j}}+6\hat{\text{k}}$ and $\vec{\text{b}}_2=-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}.$
Now,
$\vec{\text{a}}_2-\vec{\text{a}}_1=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-1&1\\-3&2& 4\end{vmatrix}$
$=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-6)^2+(-15)^2+3^2}$
$=\sqrt{36+225+9}$
$=\sqrt{270}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big).\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big)$
$=36+225+9$
$=270$
The shortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{270}{\sqrt{270}}\Big|$
$=\sqrt{270}$
$=3\sqrt{30}$

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MCQ 61 Mark

The straigth line $\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$ is:

A
parallel to $x-$axis
B
parallel to $y-$axis
C
parallel to $z-$axis
✓
perpendicular to $z-$axis

Answer

Correct option: D.

perpendicular to $z-$axis

We have
$\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$
Also, the given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
Let $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ be parpendicular to the given line.
Now,
$3\text{x}+4\text{y}+0\text{z}=0$
It is satisfied by the coordinates of $z-$axis,
i.e. $(0, 0, 1).$
Hence, the given line is perpendicular to $z-$axis.

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MCQ 71 Mark

The lines $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ and $\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$ are:

✓
Coinicident.
B
Skew.
C
Intersecting.
D
Parallel.

Answer

Correct option: A.

Coinicident.

The equation of the given lines are
$\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}\dots(1)$
$\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$
$=\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}-3}{3}\dots(2)$
Thus, the two lines are parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and pass through the points $(0, 0, 0)$ and $(1, 2, 3).$
Now,
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)\times\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=\vec{0}$
$\big[\because\vec{\text{a}}\times\vec{\text{a}}=\vec{0}\big]$

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MCQ 81 Mark

The equation of the line passing through the points $\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$ and $\text{b}\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ is:

A
$\vec{\text{r}}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\lambda\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
B
$\vec{\text{r}}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)-\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
✓
$\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3$
$(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
D
$\text{None of these}$

Answer

Correct option: C.

$\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3$
$(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$

Equation of the line passing through the points having position vectors
$\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$ and $\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ is:
$\vec{r}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big\{\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)-\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)\big\},$
where $t$ is a parameter
$=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)-\text{t}\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
$=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$

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MCQ 91 Mark

If a line makes angle $\alpha,\beta$ and $\gamma$ with the axes respectively, then $\cos2\alpha+\cos2\beta+\cos2\gamma=$

A
$-2$
✓
$-1$
C
$1$
D
$2$

Answer

Correct option: B.

$-1$

If a line makes angles $\alpha,\beta$ and $\gamma$ with the axes, then
$\cos2\alpha+\cos2\beta+\cos2\gamma=1\dots(1)$
We have
$\cos2\alpha+\cos2\beta+\cos2\gamma$
$=2\cos^2\alpha-1+2\cos^2\beta-1+2\cos^2\gamma-1$ $\big[\because\cos2\theta=2\cos^2\theta-1\big]$
$=2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)-3 [$From $(1)]$
$=2(1)-3$
$=-1$

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MCQ 101 Mark

The lines $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ and $\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$ are:

A
Parallel.
B
Intersecting.
C
Skew.
✓
Coincident.

Answer

Correct option: D.

Coincident.

The equations of the given lines are
$\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}\dots(1)$
$\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$
$\Rightarrow\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}-3}{3}\dots(2)$
Thus, the two lines are parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and pass through the points $(0, 0, 0)$ and $(1, 2, 3).$
Now,
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)\times\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=\vec{0}$ $\big[\because\vec{\text{a}}\times\vec{\text{a}}=\vec{0}\big]$
Since, the distence between the two parallel lines is $0$,
the given two lines are coincident lines.
Disclaimar: The answer given in the book is incorrect.
This solution is created according to the question given in the book.

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MCQ 111 Mark

The perpendicular distance of the point $P(1, 2, 3)$ from the line $\frac{\text{x}-6}{3}=\frac{\text{y}-7}{2}=\frac{\text{z}-7}{-2}$ is:

✓
$7$
B
$5$
C
$0$
D
None of these

Answer

Correct option: A.

$7$

$\frac{\text{x}-6}{3}=\frac{\text{y}-7}{2}=\frac{\text{z}-7}{-2}$
Let point $(1, 2, 3)$ be $P$ and the point through which the line passes be $Q(6, 7, 7)$.
Also, the line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$
Now,
$\overrightarrow{\text{PQ}}=5\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
$\therefore\vec{\text{b}}\times\overrightarrow{\text{PQ}} =\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&2&-2\\5&5&4\end{vmatrix}$
$=18\hat{\text{i}}-22\hat{\text{j}}+5\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|$
$=\sqrt{18^2+(-22)^2+5^2}$
$=\sqrt{324+484+25}$
$=\sqrt{833}$
$\because\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}$
$=\frac{\sqrt{833}}{\sqrt{17}}$
$=\sqrt{49}$
$=7$

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MCQ 121 Mark

The angle between the lines $\frac{\text{x}-1}{1}=\frac{\text{y}-1}{1}=\frac{\text{z}-1}{2}$ and $\frac{\text{x}-1}{-\sqrt{3}-1}=\frac{\text{y}-1}{\sqrt{3}-1}=\frac{\text{z}-1}{4}$ is:

A
$\cos^{-1}\big(\frac{1}{65}\big)$
B
$\frac{\pi}{6}$
✓
$\frac{\pi}{3}$
D
$\frac{\pi}{4}$

Answer

Correct option: C.

$\frac{\pi}{3}$

We have
$\frac{\text{x}-1}{1}=\frac{\text{y}-1}{1}=\frac{\text{z}-1}{2}$
$\frac{\text{x}-1}{-\sqrt{3}-1}=\frac{\text{y}-1}{\sqrt{3}-1}=\frac{\text{z}-1}{4}$
The direction ratios of the given lines are proportional to $1, 1, 2$ and $-\sqrt{3}-1,\sqrt{3}-1, 4$
The given lines are parallel to vectors $\vec{\text{b}}_1=\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}_2=\big(-\sqrt{3}-1\big)\hat{\text{i}}+\big(\sqrt{3}-1\big)\hat{\text{j}}+4\hat{\text{k}}$
Let $\theta$ be the angle between the given lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big).\big\{\big(-\sqrt{3}-1\big)\hat{\text{i}}+\big(\sqrt{3}-1\big)\hat{\text{j}}+4\hat{\text{k}}\big\}}{\sqrt{1^2+1^2+1^2}\sqrt{\big(-\sqrt{3}-1\big)^2+\big(\sqrt{3}-1\big)^2+4^2}}$
$=\frac{-\sqrt{3}-1+\sqrt{3}-1+8}{\sqrt{3}\sqrt{24}}$
$=\frac{6}{6\sqrt{2}}$
$\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\frac{\pi}{3}$

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MCQ 131 Mark

If the diraction ratios of a line are proportional to $1, -3, 2,$ then its diraction cosines are:

✓
$\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
B
$\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$
C
$-\frac{1}{\sqrt{14}},\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
D
$-\frac{1}{\sqrt{14}},-\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}}$

Answer

Correct option: A.

$\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$

The diraction ratios of the line are proportional to $1, -3, 2.$
$\therefore$ The direction cosines of the line are
$\frac{1}{\sqrt{1^2+(-3)^2+2^2}},\frac{-3}{\sqrt{1^2+(-3)^2+2^2}},\frac{2}{\sqrt{1^2+(-3)^2+2^2}}$
$=\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$

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MCQ 141 Mark

The direction ratios of the line perprndicular to the lines $\frac{\text{x}-7}{2}=\frac{\text{y}+17}{-3}=\frac{\text{z}-6}{1}$ and, $\frac{\text{x}+5}{1}=\frac{\text{y}+3}{2}=\frac{\text{z}-4}{-2}$ are proportional to:

✓
$4, 5, 7$
B
$4, -5, 7$
C
$4, -5, -7$
D
$-4, 5, 7$

Answer

Correct option: A.

$4, 5, 7$

We have
$\frac{\text{x}-7}{2}=\frac{\text{y}+17}{-3}=\frac{\text{z}-6}{1}$
$\frac{\text{x}+5}{1}=\frac{\text{y}+3}{2}=\frac{\text{z}-4}{-2}$
The direction ratios of the given lines are proportional to $2, -3, 1$ and $1, 2, -2.$
The vectors parallel to the given vectors are $\vec{\text{b}}_1=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$
Vector perpendicular to the given two lines is
$\vec{\text{b}}=\vec{\text{b}}_1\times\vec{\text{b}}_2$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-3&1\\1&2&-2\end{vmatrix}$
$=4\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}$
Hence, the direction ration of the line perpendicular to the given two lines are proportional to $4, 5, 7.$

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