Maths M.C.Q (1 Marks)

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

2. $4$

Solution:

Equation of time passing through (-1, 3, 2) and (5, 0, 6)

$\frac{\text{x}+1}{5-(-1)}=\frac{\text{y}-3}{0-3}=\frac{\text{z}-2}{6-2}$

$\frac{\text{x}+1}{6}=\frac{\text{y}-3}{-3}$

$=\frac{\text{z}-2}{4}=\text{k}$

Any point on it,

$\text{P}(6\text{k}-1,\text{3}\text{k}+3,4\text{k}+2)$

x Coordinate $=2$

$=6\text{k}-1$

$\Rightarrow\text{k}=\text{y}_2$

z Coordinate $=4\text{k}+2$

$=4\Big(\frac{1}{2}\Big)+2$

$=2+2=4$

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

4. none of these

Solution:

Suppose, l, m, n are direction cosines

⟹ 1² + m² + n² = 1

But 1 = m = n = 1

⟹ 3m² = 1

⟹ 1 = m = n = $\frac{1}{\sqrt3}$​

which are not direction cosines of either of the three co-ordinate axes.

2. No

Solution:

No, they can not be the direction cosines of any directed line.

As the sum of square of them is not 1.

$\text{As}\Big(\frac{1}{\sqrt{3}}\Big)^2+\Big(\frac{2}{\sqrt{3}}\Big)^2+\Big(\frac{-2}{\sqrt{3}}\Big)^2$

$=\frac{1+4+4}{3}$

$=3$

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,

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

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

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

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

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. 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.

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

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

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

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

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

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).

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

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

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

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

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

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,

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

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

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

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

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

$\sqrt{1+4+4}$

$=3\text{ units}$

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. < 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)

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 $\frac{12}{13},\frac{4}{13},\frac{1}{13}$

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).

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

Solution:

Area $=\frac{1}{2}|1(5-\text{b})+3(\text{b}-2)+0(2-5)|$

As points are collinear, so area = 0

$\therefore\frac{1}{2}|1(5-\text{b})+3(\text{b}-2)+0(2-\text{5})|=0$

⇒ 5 − b + 3b − 6 = 0

⇒ = 1 = 2b

$\therefore\text{b}=\frac{1}{2}$

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,

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

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

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

$\Rightarrow\theta=90^\circ$

3. 0, 1, 0

Solution:

When P moves then x = 0, z = 0 but y is not given. Let y = y Then the coordinates of the point will be (0, y, 0) Now, direction cosines with respect to (0, y, 0) is given by.

$\cos\alpha=\frac{0}{0^2+\text{y}^2+0^2}=\frac{0}{\text{y}}=0$

$\cos\beta=\frac{\text{y}}{0^2+\text{y}^2+0^2}=\frac{\text{y}}{\text{y}}=1$

$\cos\gamma=\frac{{0}}{0^2+\text{y}^2+0^2}=\frac{{0}}{\text{y}}=0$

The direction cosines are 0, 1, 0

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

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

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

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

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

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

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

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

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

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

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

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

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) $0,\frac{\sqrt{3}}{2},\frac{1}{2}$

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

2. a, 1, c

Solution:

Given x = ay + b and z = cy + d

$\Rightarrow\frac{\text{x}-\text{b}}{\text{a}}=\text{y}$​ and $\frac{\text{z}-\text{d}}{\text{c}}=\text{y}$

$\Rightarrow\frac{\text{x}-\text{b}}{\text{a}}=\frac{\text{y}}{1}=\frac{\text{z}-\text{d}}{c}$

Therefore Drs of given line is a, 1, c