Maths M.C.Q (1 Marks)

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 $

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

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

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

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

4. $3\sqrt{30}$

Solution:

We have

$\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}\dots(1)$

$\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}\dots(2)$

We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3, -1, 1.

Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1,$ where $\vec{\text{a}}_1=3\hat{\text{i}}+8\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{b}}_1=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}.$

Also, line (2) passes through the point (3, -7, 6) and has direction ratios proprtional to -3, 2, 4.

Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2,$ where $\vec{\text{a}}_2=-3\hat{\text{i}}-7\hat{\text{j}}+6\hat{\text{k}}$ and $\vec{\text{b}}_2=-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}.$

Now,

$\vec{\text{a}}_2-\vec{\text{a}}_1=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$

$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-1&1\\-3&2&4\end{vmatrix}$

$=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$

$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-6)^2+(-15)^2+3^2}$

$=\sqrt{36+225+9}$

$=\sqrt{270}$

$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big).\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big)$

$=36+225+9$

$=270$

The shortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by

$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$

$=\Big|\frac{270}{\sqrt{270}}\Big|$

$=\sqrt{270}$

$=3\sqrt{30}$

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

Solution:

x = 12, y = 4, z = 3

Direction cosines = 

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

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

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:

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

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

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

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

$\cos(\gamma)=0$

$\gamma=90^\circ$

4. $\text{x}_{2} - \text{x}_{1}, \text{y}_{2} - \text{y}_{1}, \text{z}_{2} -\text{ z}_{1}$

2. -1

Solution:

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

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

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

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

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

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

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. 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.

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

2. Parallel.

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

3. $\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$

Solution:

Equation of the line passing through the points having position vectors

$\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$ and $\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ is:

$\vec{r}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big\{\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)-\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)\big\},$ where t is a parameter

$=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)-\text{t}\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$

$=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$

1. $\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$

Solution

L makes an angle $\frac{\pi}{4}$ with X and Y axis and $\frac{\pi}{2}$

$\therefore$ d.cs are $\Big(\cos\frac{\pi}{34},\cos\frac{\pi}{4},\cos\frac{\pi}{2}\Big)=\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$

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

Solution:

Consider the problem 

Let l, m, n are direction cosines of the given line.

then as it made an acute angle with x−axis,

Therefore, l > 0

The line passes through (6, −7, −1) and (2, −3, 1)

Therefore, its direction ratios are 

6 − 2, −7 + 3, −1−1 or 2, −2, −1

Hence direction cosines of the line are given by $\frac{2}{3},\frac{2}{3},-\frac{1}{3}.$

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

Solution:

P(1, -2, 4), Q(-1, 1, -2)

$\text{PQ}=\sqrt{(1-(1))^2 +(2-1)^2+(4-(-2))^{2}}$

$=\sqrt{4+9+36}$

$=\sqrt{49}=7\text{DC}$

$=\Big(\frac{-1-1}{7},\frac{1-(2)}{7},\frac{-2-4}{7}\Big)$

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

3. 5 units

Solution:

The length of the perpendicular drawn from the point (4, -7, 3) on the y-axis is

⇒ Point on the y-axis would be = (0, -7, 0)

The length of the perpendicular drawn $=\sqrt{(4-0)^2}+(-7-(-7))^2+(3-0)^2$

$=\sqrt{4^2}+0^2+3^2$

$\Rightarrow\sqrt{16}+0+9$

$=\sqrt{25}$

$=5$

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,

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

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

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

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

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

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

1. (bl - am)y + (cl - an)z + dl - ap = 0

Solution:

The equation of the plane passing through the intersection of the planes

ax + by + cz + d = 0

and lx + my + nz + p =0

Will be $(\text{ax} + \text{by} +\text{cz} +\text{d})+\lambda(\text{lm}+\text{my}+\text{nz}+\text{p})=0$

$\text{x}(\text{a}+\lambda1)+\text{y}(\text{b}+\lambda\text{m})+\text{z}(\text{c}+\lambda\text{n})+(\text{d}+\lambda\text{p})=0\ (1)$

Since the plane is parallel to the line y = 0 and z = 0

$\text{a}+\lambda1=0$

$\lambda=\frac{-\text{a}}{\text{l}}$

Putting the value of A in eqution (1), we get

$\text{x}\Big(\text{a}+\Big(\frac{\text{-a}}{\text{l}}\Big)\text{l}\Big)+\text{y}\Big(\text{b}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{m}+\text{y}\Big(\text{c}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{n}+\text{d}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{p}=0$

$\text{y}(\text{bl}-\text{am})+\text{z}(\text{cl}-\text{an})+\text{dl}-\text{ap}=0$

Heance, option (a)

4. None of these

Solution:

 $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}-0}{3}$ and $\frac{\text{x}}{-2}=\frac{\text{y}-2}{-3}=\frac{\text{z}+1}{-1}$

Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector.

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

3. $\frac{2\sqrt{3}}{5}$

3. $\sqrt{3}$

Solution:

D.C of the line are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Any point on the line at a distance tt from P(2, -1, 2) is

$\Big(2+\frac{\text{t}}{\sqrt{3}},-1+\frac{\text{t}}{\sqrt{3}},2+\frac{\text{t}}{\sqrt{3}}\Big)$

which lies on $2\text{x} + \text{y + z} = 9$

$\Rightarrow\text{t}=\sqrt{3}$

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

4. $\frac{40}{49}$

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

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

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

1. 2x - y = 0 and y- 3z = 0

Solution:

The given plane is

$2\text{x}-(1-\lambda)\text{y}+3\lambda\text{z}=0$

$\Rightarrow(2\text{x}-\text{y})+\lambda(-\text{y}+3\text{z})=0$

So, this plane passes through the intersection of the planes

2x - y = 0 and -y + 3z = 0

⇒ 2x - y = 0 and y - 3z = 0.

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:

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

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

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

4. 0, 1, 0

Solution:

Given, x = z = 0

It represents Z-axis

$\therefore$ Direction cosines = (0, 1, 0)

4. $\frac{\sqrt{2}}{10}$

Solution:

We have, the equation of line as

$\frac{\text{x}-2}{3}=\frac{\text{y}-2}{3}=\frac{\text{z}-2}{3}$

This line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$

Equation of plane is $2\text{x}-2\text{y}+\text{z}=5$

Normal to the plane $\vec{\text{n}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$

Its angle between line and plane is $'\theta'.$

Then $\sin\theta=\frac{|\vec{\text{b}}\cdot{\vec{\text{b}}}|}{|{\vec{\text{b}}}||{\vec{\text{b}}}|}$

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

$=\frac{|6-8+5|}{\sqrt{50}\sqrt{9}}$

$=\frac{3}{15\sqrt{2}}=\frac{1}{5\sqrt{2}}$

$\sin\theta=\frac{\sqrt{2}}{10}$

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

4. perpendicular to z-axis

Solution:

We have

$\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$

Also, the given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$

Let $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ be parpendicular to the given line.

Now,

$3\text{x}+4\text{y}+0\text{z}=0$

It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).

Hence, the given line is perpendicular to z-axis.