M.C.Q (1 Marks)

MCQ 11 Mark

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

A
$\frac{3}{\sqrt{70}}, \frac{2}{\sqrt{70}}, \frac{8}{\sqrt{70}}$
✓
$\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}$
C
$\frac{2}{\sqrt{77}}, \frac{-3}{\sqrt{77}}, \frac{8}{\sqrt{77}}$
D
$\frac{8}{\sqrt{13}}, \frac{-2}{\sqrt{13}}, \frac{3}{\sqrt{13}}$.

Answer

Correct option: B.

$\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}$

(B) $\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}$

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MCQ 21 Mark

The direction cosine of $y$-axis is:

A
$0,0,0$
B
$1,0,0$
✓
$0,1,0$
D
$0,0,1$.

Answer

Correct option: C.

$0,1,0$

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MCQ 31 Mark

The distance of the point $(\alpha, \beta, \gamma)$ from $y$-axis is :

A
$\alpha$
B
$\beta$
C
$\gamma$
✓
$\sqrt{\alpha^2+\gamma^2}$

Answer

Correct option: D.

$\sqrt{\alpha^2+\gamma^2}$

(D) $\sqrt{\alpha^2+\gamma^2}$

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MCQ 41 Mark

If a line makes angles of $90^{\circ}, 135^{\circ}, 45^{\circ}$ with $x, y$ and $z$ axes, then its direction cosines will be :

A
$\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, 0$
✓
$0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$
C
$\frac{1}{2},-\frac{1}{2}, 0$
D
$0, \frac{1}{2},-\frac{1}{2}$

Answer

Correct option: B.

$0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$

$l=\cos 90^{\circ}, m=\cos 135^{\circ}, n=\cos 45^{\circ}$
$\Rightarrow l=0,$
$m=-\frac{1}{\sqrt{2}},$
$n=\frac{1}{\sqrt{2}}$
Hence direction cosines of the line are $0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.
Hence the correct option is $(B).$

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MCQ 51 Mark

If the lengths of projections of a line segment on the axes be respectively $3,4,12$, then the length of the line segment is:

A
3
B
4
C
12
✓
13

Answer

Correct option: D.

(D)
Projection on $x$-axis $=r l$
Projection on $y$-axis $=r m$
Projection on $z$-axis $=r n$
According to the question,
$
\begin{aligned}
3 & =r . l \quad \quad \ldots \ldots(1)\\
4 & =r . m \quad \quad \ldots \ldots(2)\\
12 & =r . n\quad \quad \ldots \ldots(3)
\end{aligned}
$
Squaring equations (1), (2) and (3) and adding
$
\begin{aligned}
3^2+4^2+12^2= & r^2\left(l^2+m^2+n^2\right) \\
r^2= & 9+16+144=169 \\
& \because l^2+m^2+n^2=1 \\
r^2= & \sqrt{169}=13
\end{aligned}
$
Hence the correct option is (D).

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MCQ 61 Mark

Perpendicular distance of the point $P (x, y, z)$ from $z$ axis is :

A
$\frac{z}{\sqrt{x^2+y^2+z^2}}$
✓
$\sqrt{x^2+y^2}$
C
$1$
D
$Z$

Answer

Correct option: B.

$\sqrt{x^2+y^2}$

On $z$-axis, $x=0, y=0$.
Hence a point on $z-$axis will be $(0,0, z)$.
Hence the perpendicular distance from $z-$axis
$=\sqrt{(x-0)^2+(y-0)^2+(z-z)^2}$
$=\sqrt{x^2+y^2}$
Hence the correct option is $(B).$

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MCQ 71 Mark

If the direction ratios of the lines are $a_1, b_1, c_1$ and $a_2$, $b_2, c_2$, then they will be mutually perpendicular if :

A
$a_1 b_2+a_2 b_1+c_1 c_2=0$
B
$a_1^2+b_1^2+c_1^2=1$
C
$a_2^2+b_2^2+c_2^2=1$
✓
$a_1 a_2+b_1 b_2+c_1 c_2=0$

Answer

Correct option: D.

$a_1 a_2+b_1 b_2+c_1 c_2=0$

(D)
We know that $\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$
If $\theta=90^{\circ}$ then $\cos \theta=\cos 90^{\circ}=0$
Hence putting the value, we get $a_1 a_2+b_1 b_2+c_1 c_2$ $=0$
Hence the correct option is (D).

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MCQ 81 Mark

Which of the following groups are not the direction cosines of a line?

✓
$1,1,1$
B
$0,0,-1$
C
$-1,0,0$
D
$0,-1,0$

Answer

Correct option: A.

$1,1,1$

(A)
We know that the sum of the squares of the direction cosines of any line is always 1 . Hence the group (A) is not the direction cosines of a line. Hence the correct option is (A).

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Maths M.C.Q (1 Marks)

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