Maths M.C.Q (1 Marks)

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

4. $\sqrt{\alpha^2+\gamma^2}$

2. $\underline{+}\frac{1}{\sqrt{3}}$

4. $2$

1. x - 4y + 2z + 4 = 0

Solution:

Let a, b, c be the dirction ratios of the required plane.

The given line equation can be rewritten as

$\frac{\text{x}-1}{1}=\frac{\text{y}-\frac{5}{2}}{\frac{1}{2}}=\frac{\text{z}-0}{\frac{1}{2}}\ .....(1)$

$\frac{\text{x}-0}{\frac{1}{3}}=\frac{\text{y}-\frac{11}{4}}{\frac{1}{4}}=\frac{\text{z}-\frac{4}{3}}{\frac{1}{3}}\ .....(2)$

Since the required plane is parallel to the lines (1) and (2),

$\text{a}+\frac{\text{b}}{2}+\frac{\text{c}}{2}=0\Rightarrow2\text{a}+\text{b}+\text{c}=0....(3)$

$\frac{\text{a}}{1}=\frac{\text{b}}{-4}=\frac{\text{c}}{2}=0\Rightarrow4\text{a}+3\text{b}+4\text{c}=0....(4)$

Solving (3) and (4) using cross-multiplication method, we get

$\frac{\text{a}}{1}=\frac{\text{b}}{-4}=\frac{\text{c}}{2}=\lambda\text{(say)}$

$\Rightarrow\text{a}=\lambda,\text{b}=-4\lambda,\text{c}=2\lambda$

Now, the eqution of the plane whose direction ratios are $\lambda,-4\lambda,2\lambda$ and passing through the point.

$\lambda(\text{x}-2)+(-4\lambda)(\text{y}-3)+2\lambda(\text{z}-3)=0$

$\Rightarrow\text{x}-4\text{y}+2\text{z}+4=0$

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

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)

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

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

2. $\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$

Solution:

Given,

a = (-1, 5, 4)

b = (0, 0, 1) [$\therefore$ 1 to plone z]

We know that,

$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$

$\vec{\text{r}}=(-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})+\lambda\hat{\text{k}}$

$\Rightarrow\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$

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

Solution:

Since, direction cosines of a line are k, k, and k.

$\therefore$ l = k, m = k and n = k

We know that, l² + m² + n² = 1

⇒ k² + k² + k² = 1

$\text{k}^2=\frac{1}{3}$

$\therefore\text{k}=\pm\frac{1}{\sqrt{3}}$

1. < br > (0, 1, 1) < br >

Solution:

The plane x = 2 is perpendicular to x axis So the angle is $\frac{\pi}{2},\cos\frac{\pi}{2}=0$

0 The plane x = 2 is parallel to both y axis and z axis So the angle is (0, 1, 1)

2. $\frac{10}{3\sqrt{3}}$

Solution:

The given line passes through the point whose position vector is $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$

We know that the perpendicular distance of a point P of position vector $\vec{\text{a}}$ from the plane $\vec{\text{r}}.\vec{\text{n}}=\text{d}$ is given by

$\text{P}=\frac{\big|\vec{\text{a}}.\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$

Here, $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{n}}=\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}},\text{d}=5$

So, the required distance P is given by

$\text{P}=\frac{\Big|\big(2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big),\big(\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}}\big)-5\Big|}{\Big|\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}}\Big|}$

$=\frac{|2-10+3-5|}{\sqrt{1+25+1}}$

$=\frac{|-10|}{\sqrt{27}}$

$=\frac{10}{3\sqrt{3}}\text{units}$

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

Solution:

If l, m, n are the directions cosine of a line then l² + m² + n² = 1 Thus we get $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

1. $13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$

Solution:

If a line makes angles $\alpha,\beta$ and $\gamma$ with the axes, then $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\dots(1)$

Let r be the length of the line segment. then,

$\text{r}\cos\alpha=12,\text{r}\cos\beta=4,+\cos\gamma=3\dots(2)$

$\Rightarrow\big(\text{r}\cos\alpha\big)^2+\big(\text{r}\cos\beta\big)^2+\big(\text{r}\cos\gamma\big)^2=12^2+4^3+3^2$

$\Rightarrow\text{r}^2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)=169$

$\Rightarrow\text{r}^2(1)=169$ [From (1)]

$\Rightarrow\text{r}=\sqrt{169}$

$\Rightarrow\text{r}=\pm13$

$\Rightarrow\text{r}=13$ (Since length cannot be negative)

(Since legth cannot be negative)

Substituting r = 13 in (2), we get

$\cos\alpha=\frac{12}{13},\cos\beta\frac{4}{13},\cos\gamma=\frac{1}{13}$ Thus, the direction cosines of the line are $\frac{12}{13},\frac{4}{13},\frac{1}{13}$

1. $\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}})=7$

Solution:

We know that the equation $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ represents a plane passing through a point whose position vectors is $\vec{\text{a}}$ and parrallel to the vectors $\vec{\text{b}}$ and $\vec{\text{c}}$.

Here, $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}};\ \vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}};\ \vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$

Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$

$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&1\\1&-2&3\end{vmatrix}$

$=5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$

The vector equation of the plane in scalar product from is

$\vec{\text{r}}.\vec{\text{n}}=\vec{\text{a}}.\vec{\text{n}}$

$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=\big(\hat{\text{i}}-\hat{\text{j}}-0\hat{\text{k}}\big).\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)$

$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=5+2+0$

$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=7$

$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=7$

2. Non collinear

Solution:

Area of triangle formed by these vertices is,

$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\2&3&1\\8&11&1\end{vmatrix}$

Applying R₂​ → R₂​ − R₁​, R_3​ → R_{3 ​}− R1​

$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\1&2&0\\7&10&0\end{vmatrix}=\frac{1}{2}(10-14)=2$

Hence points are non collinear

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.

4. $(\alpha,\beta,-\gamma)$

Solution:

In XY-plane, only the sign of z coordinate of the point got changed after the reflection. Therefore, the reflection of the point $(\alpha,\beta,\gamma)$ is $(\alpha,\beta,-\gamma).$

3. $1, 2, 3$

Solution:

6x - 2 = 3y + 1 = 2x - 2

$6\big(\text{x}-\frac{2}{6}\big)=3(\text{y}+\frac{1}{3}\big)=2(\text{x}-\frac{2}{2}\big)$

$\frac{\big(\text{x}-\frac{1}{3}\big)}{1}=\frac{\big(\text{y}+\frac{1}{3}\big)}{2}=\frac{(\text{x}-1)}{3}$

Line will be passing through the poits, $\Big(\frac{1}{3},-\frac{1}{3},1\Big)$

and parallel to the line having direction ratios is 1, 2, 3

1. 5x + y − 11 = 0

Solution:

Since, the points A(1, 6), B(3, −4) and C(x, y) are colinear

$\therefore\begin{vmatrix}1&6&1\\3&-4&1\\\text{x}&\text{y}&1\end{vmatrix}=0$

⇒ 1(−4−y) −6(3 − x) + 1(3y + 4x) = 0

⇒ 10x + 2y − 22 = 0

⇒ 5x + y − 11 = 0

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. 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.

1. $\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$

Solution:

Let the direction ratio of the required plane be proportinal to a, b, c.

Scince the required plane contains the line $\vec{\text{r}}=(-2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})+\lambda(3\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$

It must pass through the point (-2, -3, 4) and it should be parallel to the line.

So, the equation of the plane is

a(x + 2) + b(y + 3) + c(z - 4) = 0 ....(1) and

3a - 2b - c = 0 ....(2)

It is given that plane (1) passes through the point $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ or (1, 2, 3).

a(1 + 2) + b(2 + 3) + c(3 - 4) = 0

3a + 5b - c = 0 .......(3)

So,

Solving (1) (2) and (3), we get

$\begin{vmatrix}\text{x}+2&\text{y}+3&\text{z}-4\\3&-2&-1\\3&5&-1\end{vmatrix}=0$

$\Rightarrow7(\text{x}+2)+0(\text{y}+3)+21(\text{y}-4)=0$

$\Rightarrow\text{x}+2+3\text{z}-12=0$

$\Rightarrow\text{x}+3\text{z}=10$ or $\vec{\text{r}}.\big(\hat{\text{i}}+3\hat{\text{k}}\big)=10$

1. x - 5y + 3z = 7

Solution:

The equation of the plane passing though the line of intersection of the given planes is

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

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

This plane is parallel to the line $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}.$

It means that this line is perpendicular to the normal of the plane (1).

$\Rightarrow1(1+2\lambda)\text{x}+2(1-\lambda)+3(1+3\lambda)=0$(Because a₁a₂ + b₁b₂ + c₁c₂ = 0)

$\Rightarrow1+2\lambda+2-2\lambda+3+9\lambda=0$

$\Rightarrow9\lambda+6=0$

$\Rightarrow\lambda=\frac{-2}{3}$

Substituting this in (1), we get

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

⇒ -x + 5y - 3z + 7 = 0

⇒ x - 5y + 3z = 7

3. $2\hat{\text{i}}+7\hat{\text{j}}-13\hat{\text{k}}$

Solution:

Let the required vector be a $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\ ....(1)$

Since the vector is parallel to the line of intersection of the given planes,

3a - b + c = 0 .....(2)

a + 4b - 2c = 0 ....(3)

Solving (2) and (3), we get

$\frac{\text{a}}{-2}=\frac{\text{b}}{7}=\frac{\text{c}}{13}$

Substituting these values in (1), we get

$-2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$ which is the required vector.

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

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