MCQ

MCQ 11 Mark

The xy-plane divided the line joining the point $(-1, 3, 4)$ and $(2, -5, 6)$

A
Internally in the ratio $2 : 3$
✓
Externally in the ratio $2 : 3$
C
Internally in the ratio $3 : 2$
D
Externally in the ratio $3 : 2$

Answer

Correct option: B.

Externally in the ratio $2 : 3$

Let the $XY-$plane divide the line segment joining points
$P(-1, 3, 4)$ and $Q(2, -5, 6)$ in the ratio $k : 1.$
Using the section formula, the coordinates of the point of intersection are given by
$\Big(\frac{\text{k}(2)-1}{\text{k}+1},\frac{\text{k}(-5)+3}{\text{k}+1},\frac{\text{k}(6)+4}{\text{k}+1}\Big) $
On the $XY-$plane, the $Z-$coordinate of any point is zero.
$\Rightarrow\frac{\text{k}(6)+4}{\text{k}+1}=0$
$\Rightarrow6\text{k}+4=0$
$\Rightarrow\text{k}=\frac{-2}{3}$
Thus, the $XY-$plane divides the line segment joining the given points in the ratio $2 : 3$ externally.

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MCQ 21 Mark

For every point $P(x, y, z)$ on the $x-$axis $($except the origin$),$

A
$x = 0, y = 0, z ≠ 0$
B
$y = 0, z = 0, y ≠$ 0
✓
$y = 0, z = 0, x ≠ 0$
D
$x = y = z = 0$

Answer

Correct option: C.

$y = 0, z = 0, x ≠ 0$

Both $Y$ and $Z$ coordinates on each point of the $x-$axis are equal to zero.
The $X-$coordinate on the origin is also equal to zero.
Therefore, the $Y$ and $Z$ coordinates on each point of the $x-$axis, except the origin, are equal to zero,
While the $X-$coordinate is non$-$zero.

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MCQ 31 Mark

If $O$ is the origin, $OP = 3$ with direction ratios proportional to $-1, 2, -2$ then the coordinates of $P$ are:

✓
$(-1, 2,-2)$
B
$(1, 2, 2)$
C
$\Big(\frac{-1}{9},\frac{2}{9},\frac{-2}{9}\Big)$
D
$(3,6,-9)$

Answer

Correct option: A.

$(-1, 2,-2)$

Let the coordinates of $P$ be $(x, y, z).$ Then,
Direction ratios of $OP =$ Coordinates of $P-$Coordinates of $O-1, 2, 2 = (x - 0), (y - 0), (z - 0)$
Thus, coordinates of $P$ are $(-1, 2, -2).$

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MCQ 41 Mark

$A(3, 2, 0), B(5, 3, 2)$ and $C(-9, 6, -3)$ are the vertices of a tringle $\text{ABC.}$ if the bisector of $\angle\text{ABC}$ meets $BC$ at $D,$ then coordinates of $D$ are:

✓
$\Big(\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big)$
B
$\Big(-\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big)$
C
$\Big(\frac{19}{8},-\frac{57}{16},\frac{17}{16}\Big)$
D
$\text{none of these}$

Answer

Correct option: A.

$\Big(\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big)$

Since the bisector of $\angle\text{ABC}$ cannot meet $\text{BC,}$
the solution of this quation is not possible.
Disclaimer: This quation is wrong,
so the solution has not been provide.

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MCQ 51 Mark

Ratio in which the $xy-$plane divided the join of $(1, 2, 3)$ and $(4, 2, 1)$ is:

A
$3 : 1$ internally
✓
$3 : 1$ externally
C
$2 : 1$ internally
D
$2 : 1$ externally

Answer

Correct option: B.

$3 : 1$ externally

Suppose the $XY-$plane divides the line segment joining the points $P(1, 2, 3)$ and $Q(4, 2, 1)$ in the ratio $k : 1.$
Using the section formula,
the coordinates of the point of intersection are given by
$\Big(\frac{\text{k}(4)+1}{\text{k}+1},\frac{\text{k}(2)+2}{\text{k}+1},\frac{\text{k}(1)+3}{\text{k}+1}\Big)$
The $Z-$coordinate of any point on the $XY-$plane is zero
$\Rightarrow\frac{\text{k}(1)+3}{\text{k}+1}=0$
$\Rightarrow\text{k}+3=0$
$\Rightarrow\text{k}=-3=\frac{-3}{1}$
Thus, the $XY-$plane divided the line segment joining the given points in the ratio $3 : 1$ externally.

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MCQ 61 Mark

A rectangular parallelopiped is formed by planes drawn through the point $(5, 7, 9)$ and $(2, 3, 7)$ parallel to the coordinate planes. The length of an edge of this rectangular parallelopiped is:

A
$2$
B
$3$
C
$4$
✓
all of these

Answer

Correct option: D.

all of these

The give point $(5, 7, 9)$ and $(2, 3, 7)$ are two diagonally opposite vertices of the parallelopiped as all of theire coordinates.
Edges of the paralleloppiped $= |5 - 2|, |7 - 3|, |9 - 7|$
$=3, 4, 2.$

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MCQ 71 Mark

The angle between the two diagonals of a cube is:

✓
$30^\circ$
B
$45^\circ$
C
$\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
D
$\cos^{-1}\Big(\frac{1}{{3}}\Big)$

Answer

Correct option: A.

$30^\circ$

Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure.
Clearly, $\text{OP, AR,}$ Consider the diagonals $\text{OP}$ and $\text{AR.}$
Direction ratios of $\text{OP}$ and $\text{AR}$ are proportional to $a - 0, a - 0, a - 0$ and $0 - a, a - 0, a - 0,$
i.e. $a, a, a$ and $-a, a, a,$ respectivelly.
Let $\theta$ be the angle between $\text{OP}$ and $\text{AR.}$ Then,
$\cos\theta=\frac{\text{a}\times-\text{a}+\text{a}\times\text{a}+\text{a}\times\text{a}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{(-\text{a})^2+\text{a}^2+\text{a}^2}}$
$\Rightarrow\cos\theta=\frac{-\text{a}+\text{a}^2+\text{a}^2}{\sqrt{3\text{a}^2}\sqrt{3\text{a}^2}}$
$\Rightarrow\cos\theta=\frac{1}{3}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{3}\Big)$
Similarly, the angles between other pairs of the diagonals are equal to $\cos^{-1}\Big(\frac{1}{3}\Big)$ as the angle between any two diagonals.

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MCQ 81 Mark

For every point $P(x, y, z)$ on the $xy-$plane,

A
$x = 0$
B
$y = 0$
✓
$z = 0$
D
$x = y = z = 0$

Answer

Correct option: C.

$z = 0$

The $Z-$coordinate of every point on the $XY-$plane is zero.

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MCQ 91 Mark

If the $x-$coordinate of a point $P$ on the join of $Q(2, 2, 1)$ and $R(5, 1, -2)$ is $4$, then its $z-$coordinate is:

A
$2$
B
$1$
✓
$-1$
D
$-2$

Answer

Correct option: C.

$-1$

Suppose the point $P$ divided the line segment joining the point $Q(2, 2, 1)$ and $R(5, 1, -2)$ in the ratio $k : 1.$
Using the section formula, the coordinates of the point of intersection are given by
$\Big(\frac{\text{k}(5)+2}{\text{k}+2},\frac{\text{k}(1)+2}{\text{k}+1},\frac{\text{k}(-2)+1}{\text{k}+1}\Big) $
On the $XY-$plane, the $Z-$coordinate of any point is zero.
$\Rightarrow\frac{\text{k}(5)+2}{\text{k}+2}=4$
$\Rightarrow5\text{k}+2=4(\text{k}+1)$
$\Rightarrow\text{k}=2$
Now,
$Z-$coordinate of $P =\frac{\text{k}(-2)+1}{\text{k}+1}$
$\frac{2(-2)+1}{2+1}\ [\text{Substituting} \text{ k}=2]$
$=-1$

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MCQ 101 Mark

If $P(3, 2, -4), Q(5, 4, -6)$ and $R(9, 8, -10)$ are collinear, then $R$ divided $PQ$ in the ratio:

A
$3 : 2$ internally
✓
$3 : 2$ externally
C
$2 : 1$ internally
D
$2 : 1$ externally

Answer

Correct option: B.

$3 : 2$ externally

Suppose the point $R$ divides $PQ$ in the ratio $\lambda:1$.
Coordinates of $R$ are $\Big(\frac{5\lambda+3}{\lambda+1},\frac{4\lambda+2}{\lambda+1},\frac{-6\lambda-4}{\lambda+1}\Big)$.
But the coordinates of $R$ are $(9, 8, -10)$.
$\therefore\frac{5\lambda+3}{\lambda+1}=9,\frac{4\lambda+2}{\lambda+1}=8$ and $\frac{-6\lambda-4}{\lambda+1}=-10$
From each of these equations, we get
$\lambda=-\frac{3}{2}$
$\therefore$ $R$ divided $PQ$ in the ratio $3 : 2$ externally.

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MCQ 111 Mark

A parallelopiped is formed by planes drawn through the point $(2, 3, 5)$ and $(5, 9, 7)$ parallel to the coordinate planes. The length of a diagonal of the parallelopiped is:

✓
$7$
B
$\sqrt{38}$
C
$\sqrt{155}$
D
$\text{none of these}$

Answer

Correct option: A.

$7$

The given point $(2, 3, 5)$ and $(5, 9, 7)$ are two diagonally opposite vertices of the parallelopiped as all of theire coordinates are different.
$\therefore$ Edges of the paralleloppiped
$= |2 - 5|, |3 - 9|$ and $|5 - 7|$
$=3, 6$ and $2.$
Now,
Length of the diagonal of the parallelopiped
$=\sqrt{3^2+6^2+2^2}$
$=\sqrt{9+36+4}$
$=\sqrt{49}$
$=7$
Hence, length of the diagonal of the parallelepiped formed by the planes
Parallel to coordinate planes and drawn through point $(2, 3, 5)$ and $(5, 9, 7)$ is $7$ units.

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MCQ 121 Mark

The distance of the point $P(a, b, c)$ from the $x-$axis is:

✓
$\sqrt{\text{b}^2+\text{c}^2}$
B
$\sqrt{\text{a}^2+\text{c}^2}$
C
$\sqrt{\text{a}^2+\text{b}^2}$
D
$\text{none of these}$

Answer

Correct option: A.

$\sqrt{\text{b}^2+\text{c}^2}$

The projection of the point $P(a, b, c)$ on the $x-$axis is $a, (0, 0)$ as both $Y$ and $Z$ coordinates on any point on the $x-$axis are equal to zero.
$\therefore$ Distance of $P(a, b, c)$ from $x-$axis $=$ Distance of $P(a, b, c)$ from $a, (0, 0)$
$=\sqrt{(\text{a}-\text{a})^2+(\text{b}-0)^2+(\text{c}-0)^2}$
$=\sqrt{\text{b}^2+\text{c}^2}$

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MCQ 131 Mark

If a line makes angles $\alpha,\beta,\gamma,\delta$ with four diagonals of a cube, then $\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta$ is equal to:

A
$\frac{1}{3}$
B
$\frac{2}{3}$
✓
$\frac{4}{3}$
D
$\frac{8}{3}$

Answer

Correct option: C.

$\frac{4}{3}$

Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure. Clearly, $\text{OP, AR}$
The direction ratiosm of $\text{OP, AR, BS}$ and $\text{CQ}$ are
$a - 0, a - 0, a - 0,$ i.e. $a, a, a$
$0 - a, a - 0, a - 0,$ i.e. $-a, a, a$
$a - 0, 0 - a, a - 0,$ i.e. $a, -a, a$
$a - 0, a - 0, 0 - a,$ i.e. $a, a, -a$
Let the direction ratios of a line be proportional to $l, m$ and $n.$
Suppose this line makes angles $\alpha,\beta,\gamma$ and $\delta$ with $\text{OP, AR.}$
Now, $\alpha$ is the angle between $OP$ and the line whose direction ratios are proportional to $l, m$ and $n.$
$\cos\alpha=\frac{\text{a}.\text{l}+\text{a}.\text{m}+\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$\Rightarrow\cos\alpha=\frac{\text{l}+\text{m}+\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
Since $\beta$ is the angle between $AR$ and the line with direction ratios proportional to $l, m$ and $n,$ we get
$\cos\beta=\frac{-\text{a}.\text{l}+\text{a}.\text{m}+\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$\Rightarrow\cos\beta=\frac{-\text{l}+\text{m}+\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
Similarly,
$\cos\gamma=\frac{\text{a}.\text{l}-\text{a}.\text{m}+\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$\Rightarrow\cos\gamma=\frac{\text{l}-\text{m}+\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$\cos\delta=\frac{\text{a}.\text{l}+\text{a}.\text{m}-\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$\Rightarrow\cos\delta=\frac{\text{l}+\text{m}-\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta$
$=\frac{(\text{l}+\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(-\text{l}+\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(\text{l}-\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(\text{l}+\text{m}-\text{n})^2}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$=\frac{1}{3(\text{l}^2+\text{m}^2+\text{n}^2)}\Big\{(\text{l}+\text{m}+\text{n})^2+(-\text{l}+\text{m}+\text{n})^2+(\text{l}-\text{m}+\text{n})^2+(\text{l}+\text{m}-\text{n})^2\Big\}$
$=\frac{1}{3(\text{l}^2+\text{m}^2+\text{n}^2)}4\big(\text{l}^2+\text{m}^2+\text{n}^2\big)$
$=\frac{4}{3}$.

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