Maths M.C.Q (1 Marks)

1. $(-1, 2,-2)$

Solution:

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).

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

Solution:

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.

1. 7

Solution:

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.

2. 3 : 1 externally

Solution:

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.

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

Solution:

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)

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

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

3. $\frac{4}{3}$

Solution:

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,

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

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

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

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

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

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

2. Externally in the ratio 2 : 3

Solution:

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.

1. $\cos^{-1}\Big(\frac{1}{{3}}\Big)$

Solution:

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.

3. -1

Solution:

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

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

$=-1$

2. 3 : 2 externally

Solution:

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.