Maths M.C.Q (1 Marks)

4. $\frac{2}{\sqrt{29}}.$

1. $\Big(\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}}\Big)$

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

Solution:

Equation of line bisecting XOY is x = y

$\therefore$ d.r.s are (1, 1, 0)

And thus d.c.s are $\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$

2. 2

Solution:

Given expression, $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$

$=(1-\cos^2\alpha)+(1-\cos^2\beta)+(1-\cos^2\gamma)$

$=3-\cos^2\alpha+\cos^2\beta+\cos^2\gamma=3-1=2$

$(\because\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1)$

1. $3,1,-2$

Solution:

We have

x - y + z - 5 = 0 = x - 3y - 6

⇒ x - 3y - 6=0

x - y + z - 5 = 0

⇒ x = 3y + 6 .....(1)

x - y + z - 5 = 0.....(2)

From (1) and (2), we get

3y + 6 - y + z - 5 = 0

⇒ 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.

4. 2

Solution

The direction cosines of the line are

$\text{l}^2+\text{m}^2+\text{n}^2=1$

Now, $\Rightarrow\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$

$\Rightarrow1-\sin^2\alpha+1-\sin^2\beta+1-\sin^2\gamma=1$

$\Rightarrow\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2$

2. $\frac{\pi}{3}$

Solution:

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 $

$\Rightarrow\cos^2\gamma=1-\frac{3}{4}$

$\Rightarrow\cos^2\gamma=\frac{1}{4}$

$\Rightarrow\cos\gamma=\frac{1}{2}$

$\Rightarrow\gamma=\frac{\pi}{3}$

3. 11

Solution:

Let a, b, c be the projection of a line on the coordinate axes.

Then the length of the line given by $\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}$

Here we have 12² + 3² + k² = 169

$\Rightarrow\text{k}=\underline{+}4$

Thus k² - 2k + 3 = 11 or 27.

3. $\frac{1}{6}$

Solution:

Multiplying the first equation of the plane by

4x + 4y - 2z + 4 = 0

4x + 4y - 2z = -4 .....(1)

The second eqution of the plane is

4x + 4y - 2z + 5 = 0

4x + 4y - 2z = -5 .....(2)

We know that the distance between two planes ax + by + cz = d₁ and ax + by + cz = d₂ is,

$=\frac{|\text{d}_2-\text{d}_1|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$

So, the required distance

$=\frac{|-5+4|}{\sqrt{4^2+4^2+(-2)^2}}$

$=\frac{|-1|}{\sqrt{16+16+4}}$

$=\frac{1}{\sqrt{36}}$

$=\frac{1}{6}\text{units}$

2. 90°

Solution:

$\Big(\cos\frac{\pi}{12}\Big)^2+\Big(\cos\frac{5\pi}{12}\Big)^2+\big(\cos(\gamma)\big)^2=1$

$\Big(\cos\frac{\pi}{12}\Big)^2+\Big(\cos\frac{\pi}{12}\Big)^2+\big(\cos(\gamma)\big)^2=1..$

$\Big(\cos\theta=\sin\Big(\frac{\pi}{2}- \theta\Big)\Big)$

$\Big(\cos(\gamma)\Big)^2=0$

$\cos(\gamma)=0$

$\gamma=90^\circ$

2. -1

Solution:

We need to find value of $\cos2\alpha+\cos2\beta+\cos2\gamma$

It is further equal to $\cos^2\alpha-1+\cos^2\beta-1+\cos^2\gamma-1$

$=2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)-3$

$= 2(1) - 3 = 2 = -1$

$\therefore(\text{l}^2 + \text{m}^2 + \text{n}^2 = 1)$

3. $-3$

Solution:

The drs of AB are (k, 1, -2)

The drs of BC are (3, 1, -4)

Since, they are perpendicular, AB.BC = 0

3k + 1 + 8 = 0

k = -3

1. $\frac{17}{\sqrt{77}}$

2. $\frac{12}{13},\frac{4}{13},\frac{3}{13}$

Solution:

x = 12, y = 4, z = 3

Direction cosines = 

$\frac{\text{x}}{\text{x}^2 + \text{y}^2+\text{z}^2},\frac{\text{y}}{\text{x}^2 + \text{y}^2+\text{z}^2},\frac{\text{x}}{\text{x}^2 + \text{y}^2+\text{z}^2}$

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

1. 4, 5, 7

Solution:

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.

1. $\frac{1}{3}$

3. $\frac{13}{\sqrt{21}}$

1. $\frac{3}{5}$

Solution:

If a line makes the angle $\alpha,\beta,\gamma$ with x, y, z axix respectively then

l² + m² + n² = 1

⇒ 2l² + m² = 1 or 2n² + m² = 1

$\Rightarrow2\cos^2\theta=1-\cos^2\beta (\alpha=\gamma=\theta)$

$2\cos^2\theta=\sin^2\beta$

$\Rightarrow2\cos^2\theta=3\sin^2\theta$

$\Rightarrow5\cos^2\theta=3$

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

2. $\frac{2}{3}, \frac{-1}{3},\frac{-2}{3}$

Solution:

Direction cosines are.

$=\frac{2}{2^2+(-1)^2+(-2)^2},\frac{1}{2^2+(-1)^2+(-2)^2},\frac{-2}{2^2+(-1)^2+(-2)^2}$

$=\frac{2}{3}, \frac{-1}{3},\frac{-2}{3}$

3. 3

Solution:

The coordinates of any point on the given line are of the from

$\frac{\text{x}-3}{\text{1}}=\frac{\text{y}-4}{\text{2}}=\frac{\text{z}-5}{\text{2}}=\lambda$

$\Rightarrow \text{x}=\lambda+3;\text{y}=2\lambda+4;\text{z}=2\lambda+5$

So, the coordinates of the point on the given line are $(\lambda+3,2\lambda+4,2\lambda+5)$

This point lies on the plane

x + y + z = 17

$\Rightarrow\lambda+3,2\lambda+4+2\lambda+5=17$

$\Rightarrow5\lambda=5$

$\Rightarrow\lambda=1$

So, the coordinates of the point are

$(\lambda+3,2\lambda+4,2\lambda+5)$

$=(1+3,2(1))+4,2(1)+5)$

$=(4,6,7)$

Now, the distance between the points (4, 6, 7) and (3, 4, 5) is

$\sqrt{(3+4)^2+(4-6)^2+(5-7)^2}$

$\sqrt{1+4+4}$

$=3\text{ units}$

4. $\text{k}=\frac{\sqrt{1}}{3}$ or $\frac{\sqrt{1}}{3}$

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.

2. $5\sqrt{2}$

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. $\Big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big)$

Solution:

We know sum of the squares of the direction cosines is one.

i.e. $\cos^2\alpha+\cos^2\gamma=1$ 

but its given that $\alpha=\beta=\gamma\therefore\cos^2\alpha=1$

$3\cos^2\alpha=1$

$\therefore\cos^2\alpha=\frac{1}{3}$

$\therefore$ Positive directions of the axes are $\Big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big)$

3. (-3, 5, 2)

Solution:

Let Q be the image of the point P(1, 3, 4) in the plane 2x - y +z + 3 = 0

Then PQ is normal to the plane.

So, the direction ratios of PQ are proportional to 2, -1, 1 equation of PQ is

Let the coordinates of Q be (2r + 1, -r + 3, r + 4)

Let R be the mid point of PQ.

Then,

$\text{R}=\Big(\frac{2\text{r}+1+1}{2},\frac{-\text{r}+3+3}{2},\frac{\text{r}+4+4}{2}\Big)$

$=\Big(\text{r}+1,\frac{-\text{r}+6}{2},\frac{\text{r}+8}{2}\Big)$

Since R lies in the plane 2x - y + z + 3 = 0,

$2(\text{r}+1)-\Big(\frac{-\text{r}+6}{2}\Big)+\frac{\text{r}+8}{2}+3=0$

⇒ 4r + 4 + r - 6 + r + 8 + 6 = 0

⇒ 6r + 12 = 0

⇒ r = -2

Substituting this in the coordinates of Q, we get

Q = (2r + 1, -r + 3, r + 4)

=(2 (-2) + 1, 2 + 3, -2 + 4)

=(-3, 5, 2).

4. 90°

Solution:

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,

$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$

$=\frac{\big(2\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)}{\sqrt{2^2+5^2+4^2}\sqrt{1^2+2^2+(-3)^2}}$

$=\frac{2+10-12}{\sqrt{45}\sqrt{14}}$

$\Rightarrow\theta=90^\circ$

4. $\frac{-2}{7},\frac{3}{7},\frac{-6}{7}$

Solution:

Given the points are P(1, -2, 4) and Q(-1, 1, -2) Now the direction

ratios of the ray PQ are (-1 - 1, 1 + 2, -2 - 4) = (-2, 3, -6)

The direction cosines of the line PQ will be

$\Big(\frac{2}{\sqrt{2^2+3^2+6^2}},\frac{3}{\sqrt{2^2+3^2+6^2}},\frac{-6}{\sqrt{2^2+3^2+6^2}}\Big)=\Big(\frac{-2}{7},\frac{3}{7},\frac{-6}{7}\Big)$

3. $0,\frac{\sqrt{3}}{2},\frac{1}{2}$

Solution:

If direction cosine of a line is l, m, n then

l² + m² + n² = 1

$=0^2+\Big(\frac{\sqrt{3}}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2=1$

The correct answer from the given alternative is (c) $0,\frac{\sqrt{3}}{2},\frac{1}{2}$

3. (3, 2, 1)

Solution:

Equations of the planes are x = 3z + 4 and y = 2z - 3

$\therefore$ The equation of the plane passing through the line of intersection of these planes is x = 3z + 4 and y = 2z - 3

Thus The direction Ratios of the equation passes through intersection of the planes is (3, 2, 1).

3. $\frac{1}{3\sqrt{3}}$

Solution:

$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma)=1$

$\Rightarrow3\cos^2(\alpha)=1$

$\Rightarrow\cos\alpha=\underline{+}\frac{1}{\sqrt{3}}$

1. $\frac{1}{\sqrt[5]{2}}$

Solution:

Equation of plane $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$

3x + 4y + 5z − 1 = 0

diatance from origin $\frac{1}{\sqrt{150}}=\frac{1}{\sqrt[5]{2}}$

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

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

$=-1$