M.C.Q (1 Marks)

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.

View full question & answer→

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.

View full question & answer→

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).

View full question & answer→

MCQ 41 Mark

A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3) are the vertices of a tringle 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 BC, the solution of this quation is not possible.

Disclaimer: This quation is wrong, so the solution has not been provide.

View full question & answer→

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.

View full question & answer→

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.

View full question & answer→

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, OP, AR, Consider the diagonals OP and AR.
Direction ratios of OP and AR are proportional to a - 0, a - 0, a - 0 and 0 - a, a - 0, a - 0, e.i. a, a, a and -a, a, a, respectivelly.
Let $\theta$ be the angle between OP and 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.

View full question & answer→

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.

View full question & answer→

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 $=\frac{\text{k}(-2)+1}{\text{k}+1}$

$\frac{2(-2)+1}{2+1}\ [\text{Substituting} \text{ k}=2]$

$=-1$

View full question & answer→

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.

View full question & answer→

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.

View full question & answer→

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)

$=\sqrt{(\text{a}-\text{a})^2+(\text{b}-0)^2+(\text{c}-0)^2}$

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

View full question & answer→

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, OP, AR

The direction ratiosm of OP, AR, BS and 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 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,

$\cos\gamma=\frac{\text{a}.\text{l}-\text{a}.\text{m}+\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}\Rightarrow\cos\gamma=\frac{\text{l}-\text{m}+\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$

$\cos\delta=\frac{\text{a}.\text{l}+\text{a}.\text{m}-\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}\Rightarrow\cos\delta=\frac{\text{l}+\text{m}-\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$

$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta$

$=\frac{(\text{l}+\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(-\text{l}+\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(\text{l}-\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(\text{l}+\text{m}-\text{n})^2}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$

$=\frac{1}{3(\text{l}^2+\text{m}^2+\text{n}^2)}\Big\{(\text{l}+\text{m}+\text{n})^2+(-\text{l}+\text{m}+\text{n})^2+(\text{l}-\text{m}+\text{n})^2+(\text{l}+\text{m}-\text{n})^2\Big\}$

$=\frac{1}{3(\text{l}^2+\text{m}^2+\text{n}^2)}4\big(\text{l}^2+\text{m}^2+\text{n}^2\big)=\frac{4}{3}$.

View full question & answer→

MATHS M.C.Q (1 Marks)

Test yourself on this topic