3D Geometry — Page 19 | JEE Mathematics

Q181

Let two vertices of a triangle ABC be (2, 4, 6) and (0,

-

2,

-

5), and its centroid be (2, 1,

-

1). If the image of the third vertex in the plane

x+2y+4z=11

is

(\alpha,\beta,\gamma)

, then

\alpha\beta+\beta\gamma+\gamma\alpha

is equal to :

A 72

B 74

C 76

D 70

Correct Answer

Option B

Solution

Given that two vertex of a $\triangle A B C$ be $A(2,4,6)$ and $B(0,-2,-5)$ and $G=$ centroid $=(2,1,-1)$ [Given] Let, the other vertex is $C(x, y, z)$ According to the question,

\begin{aligned} & \frac{2+0+x}{3}=2 \\\\ &\Rightarrow x =4 \end{aligned}

\begin{aligned} &\frac{4-2+y}{3}=1 \\\\ &\Rightarrow y=1 \end{aligned}

\begin{aligned} & \frac{6-5+z}{3}=-1 \\\\ &\Rightarrow z=1 \end{aligned}

Hence, third vertex is $C(4,1,-4)$ Now, if image of $C(4,1,-4)$ in the plane $x+2 y+4 z=11$ is $D(\alpha, \beta, \gamma)$ .

So, $\dfrac{\alpha-4}{1}=\dfrac{\beta-1}{2}=\dfrac{\gamma+4}{4}=k$ (say)

\Rightarrow \alpha=k+4, \beta=2 k+1, \gamma=4 k-4

Then, co-ordinates of $D$ is $(k+4,2 k+1,4 k-4)$ Let, $E$ be the mid-point of $C D$ , which lies on the plane $x+2 y+4 z=11$ .

Then, co-ordinates of $E$ is

\left(\frac{k+8}{2}, \frac{2 k+2}{2}, \frac{4 k-8}{2}\right)=\left(\frac{k+8}{2}, k+1,2 k-4\right)

Since, $E$ lies on the plane, $x+2 y+4 z-11=0$

\begin{aligned} & \text{ So, } \frac{k+8}{2}+2 k+2+8 k-16-11=0 \\\\ & \Rightarrow \frac{k+8+4 k+4+16 k-54}{2}=0 \end{aligned}

\begin{array}{rlrl} & \Rightarrow 21 k =42 \\\\ & \Rightarrow k =2 \end{array}

Hence, $D \equiv(6,5,4) \equiv(\alpha, \beta, \gamma)$ So, $\alpha=6, \beta=5, \gamma=4$ Now, $\alpha \beta+\beta \gamma+\gamma \alpha=6 \times 5+5 \times 4+4 \times 6$ $=30+20+24=74$

Q182

Let P be the point of intersection of the line

{{x + 3} \over 3} = {{y + 2} \over 1} = {{1 - z} \over 2}

and the plane

x+y+z=2

. If the distance of the point P from the plane

3x - 4y + 12z = 32

is q, then q and 2q are the roots of the equation :

A

{x^2} + 18x - 72 = 0

B

{x^2} - 18x - 72 = 0

C

{x^2} + 18x + 72 = 0

D

{x^2} - 18x + 72 = 0

Correct Answer

Option D

Solution

Given, equation of line is

\begin{aligned} & \frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}=k \\\\ & \therefore x=3 k-3, y=k-2, z=1-2 k \end{aligned}

Since, given that $P \equiv(3 k-3, k-2,1-2 k)$ be the point of intersection of the given line and the plane $x+y+z=2$

\begin{aligned} &\text{ So, }(3 k-3)+(k-2)+(1-2 k) =2 \\\\ &\Rightarrow 2 k-4=2 \Rightarrow k =3 \end{aligned}

Thus, $P=(6,1,-5)$ Now, distance of point $P$ from the plane $3 x-4 y+12 z=32$ is

\begin{array}{rlrl} &q =\left|\frac{18-4-60-32}{\sqrt{9+16+144}}\right|=\left|\frac{-78}{\sqrt{169}}\right| \\\\ &\Rightarrow q =\left|\frac{-78}{13}\right|=\frac{78}{13}=6 \\\\ &\therefore q =6 \Rightarrow 2 q=12 \end{array}

Thus, $q$ and $2 q$ are roots of the equation $x^2-18 x+72=0$

Q183

For

\mathrm{a}, \mathrm{b} \in \mathbb{Z}

and

|\mathrm{a}-\mathrm{b}| \leq 10

, let the angle between the plane

\mathrm{P}: \mathrm{ax}+y-\mathrm{z}=\mathrm{b}

and the line

l: x-1=\mathrm{a}-y=z+1

be

\cos ^{-1}\left(\dfrac{1}{3}\right)

. If the distance of the point

(6,-6,4)

from the plane P is

3 \sqrt{6}

, then

a^{4}+b^{2}

is equal to :

A 48

B 85

C 32

D 25

Correct Answer

Option C

Solution

We have, $\theta=\cos ^{-1} \dfrac{1}{3}$

\begin{aligned} & \Rightarrow \cos \theta=\frac{1}{3} \\\\ & \therefore \sin \theta=\sqrt{1-\left(\frac{1}{3}\right)^2}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3} \end{aligned}

The given plane line and are

a x+y-z=b

x-1=a-y=z+1

\begin{aligned} & \therefore \sin \theta=\frac{a \cdot 1+(1)(-1)+(-1)(1)}{\sqrt{a^2+1^2+1^2} \sqrt{1^2+1^2+1^2}} \\\\ & \Rightarrow \frac{a-1-1}{\sqrt{a^2+2} \sqrt{3}}=\frac{2 \sqrt{2}}{3} \\\\ & \Rightarrow 3(a-2)=2 \sqrt{6} \sqrt{a^2+2} \\\\ & \Rightarrow 9\left(a^2+4-4 a\right)=24\left(a^2+2\right) \\\\ & \Rightarrow 9 a^2+36-36 a=24 a^2+48 \\\\ & \Rightarrow 15 a^2+36 a+12=0 \\\\ & \Rightarrow 5 a^2+12 a+4=0 \\\\ & \Rightarrow 5 a^2+10 a+2 a+4=0 \\\\ & \Rightarrow 5 a(a+2)+2(a+2)=0 \\\\ & \Rightarrow a=\frac{-2}{5},-2 \end{aligned}

So, $a=-2$ [

\because a \in Z

] Hence, the eqn. of plane is $-2 x+y-z-b=0$

\begin{aligned} & \text{ Now, } d=\left|\frac{-12-6-4-b}{\sqrt{4+1+1}}\right|=3 \sqrt{6} \\\\ & \Rightarrow|-(b+22)|=18 \\\\ & \Rightarrow b=18-22=-4 \\\\ & \therefore a^4+b^2=(-2)^4+(-4)^2 \\\\ & =16+16=32 \end{aligned}

Q184

Let

\mathrm{P}

be the plane passing through the line

\dfrac{x-1}{1}=\dfrac{y-2}{-3}=\dfrac{z+5}{7}

and the point

(2,4,-3)

. If the image of the point

(-1,3,4)

in the plane P is

(\alpha, \beta, \gamma)

then

\alpha+\beta+\gamma

is equal to :

A 10

B 12

C 9

D 11

Correct Answer

Option A

Solution

Equation of line : $\dfrac{x-1}{1}=\dfrac{y-2}{-3}=\dfrac{z+5}{7}$ Let $B \equiv(2,4,-3)$ So, $\overrightarrow{\mathrm{AB}}=(2-1) \hat{i}+(4-2) \hat{j}+(-3+5) \hat{k}$ $=\hat{i}+2 \hat{j}+2 \hat{k}$

\begin{aligned} & \vec{n}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 7 \\ 1 & 2 & 2 \end{array}\right|=(-6-14) \hat{i}-(2-7) \hat{j}+(2+3) \hat{k} \\\\ & =-20 \hat{i}+5 \hat{j}+5 \hat{k} \\\\ & =-5(4 \hat{i}-\hat{j}-\hat{k}) \end{aligned}

$\therefore$ Equation of plane is :

4(x-1)+(-1)(y-2)-1(z+5)=0

\begin{aligned} & \Rightarrow 4 x-4-y+2-z-5=0 \\\\ & \Rightarrow 4 x-y-z-7=0 \end{aligned}

$\because$ Image of point $(-1,3,4)$ is $(\alpha, \beta, \gamma)$

\begin{aligned} & \text{ So, } \frac{\alpha+1}{4}=\frac{\beta-3}{-1}=\frac{\gamma-4}{-1}=\frac{-2(-4-3-4-7)}{16+1+1}=2 \\\\ & \Rightarrow \alpha=7, \beta=1, \gamma=2 \\\\ & \text{ So, } \alpha+\beta+\gamma=10 \end{aligned}

Q185

The shortest distance between the lines

\dfrac{x-4}{4}=\dfrac{y+2}{5}=\dfrac{z+3}{3}

and

\dfrac{x-1}{3}=\dfrac{y-3}{4}=\dfrac{z-4}{2}

is :

A

3 \sqrt{6}

B

6 \sqrt{2}

C

6 \sqrt{3}

D

2 \sqrt{6}

Correct Answer

Option A

Solution

The given lines are

\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}

and

\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}

\begin{aligned} & \text{ So, } \vec{b}_1=4 \hat{i}+5 \hat{j}+3 \hat{k} \\\\ & \vec{b}_2=3 \hat{i}+4 \hat{j}+2 \hat{k} \\\\ & \vec{a}_1=4 \hat{i}-2 \hat{j}-3 \hat{k} \\\\ &\vec{a}_2=\hat{i}+3 \hat{j}+4 \hat{k} \end{aligned}

\begin{aligned} & \therefore \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{array}\right| \\\\ & =(10-12) \hat{i}-(8-9) \hat{j}+(16-15) \hat{k} \\\\ & =-2 \hat{i}+\hat{j}+\hat{k} \end{aligned}

\begin{aligned} & \text{ Shortest distance, } d=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right| \\\\ & =\left|\frac{(3 \hat{i}-5 \hat{j}-7 \hat{k}) \cdot(-2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{4+1+1}}\right| \\\\ & =\left|\frac{-6-5-7}{\sqrt{6}}\right|=\frac{18}{\sqrt{6}}=3 \sqrt{6} \text{ units } \end{aligned}

Q186

If the equation of the plane containing the line

x+2 y+3 z-4=0=2 x+y-z+5

and perpendicular to the plane

\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})

is

a x+b y+c z=4

, then

(a-b+c)

is equal to :

A 18

B 22

C 20

D 24

Correct Answer

Option B

Solution

Equation of plane $\mathrm{P}$ containing the given lines is

\begin{aligned} & (x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0 \\\\ & \Rightarrow(1+2 \lambda) x+(2+\lambda) y+(3-\lambda) z+(-4+5 \lambda)=0 \end{aligned}

Now, plane $\mathrm{P}$ is perpendicular to plane $\mathrm{P}^{\prime}$

\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})

So, normal to plane $\mathrm{P}^{\prime}$ is

\begin{aligned} & \vec{n}=(\hat{i}+\hat{j}+\hat{k}) \times(\hat{i}-2 \hat{j}+3 \hat{k}) \\\\ & \Rightarrow \vec{n}=5 \hat{i}-2 \hat{j}-3 \hat{k} \end{aligned}

$\therefore \mathrm{P}$ and $\mathrm{P}^{\prime}$ are perpendicular

\begin{aligned} & \therefore 5(1+2 \lambda)-2(2+\lambda)-3(3-\lambda)=0 \\\\ & \Rightarrow 5+10 \lambda-4-2 \lambda-9+3 \lambda=0 \\\\ & \Rightarrow 11 \lambda=8 \Rightarrow \lambda=\frac{8}{11} \end{aligned}

\begin{array}{r} \therefore P:\left(1+\frac{16}{11}\right) x+\left(2+\frac{8}{11}\right) y+\left(3-\frac{8}{11}\right) z+\left(5 \times \frac{8}{11}-4\right) =0 \end{array}

i.e., $27 x+30 y+25 z=4$ which is same as $a x+b y+c z=4$

\begin{aligned} & \therefore a=27, b=30 \text{ and } c=25 \\\\ & \Rightarrow a-b+c=27-30+25=22 \end{aligned}

Q187

If the equation of the plane passing through the line of intersection of the planes

2 x-y+z=3,4 x-3 y+5 z+9=0

and parallel to the line

\dfrac{x+1}{-2}=\dfrac{y+3}{4}=\dfrac{z-2}{5}

is

a x+b y+c z+6=0

, then

a+b+c

is equal to :

A 13

B 15

C 14

D 12

Correct Answer

Option C

Solution

Equation of plane intersection of two plane

\mathrm{P}_1+\lambda \mathrm{P}_2=0

\mathrm{P}_1: 2 x-y+z=3, \text{ and } \mathrm{P}_2: 4 x-3 y+5 z+9=0

Equation of any plane passing through the intersection of given planes is $(2 x-y+z-3)+\lambda(4 x-3 y+5 z+9)=0$

\Rightarrow(2+4 \lambda) x+(-1-3 \lambda) y+(1+5 \lambda) z+(-3+9 \lambda)=0

..........(i) Plane (i) is parallel to given line

\begin{aligned} & \therefore (-2)(2+4 \lambda)+4(-1-3 \lambda)+5(1+5 \lambda) =0 \\\\ & \Rightarrow -4-8 \lambda-4-12 \lambda+5+25 \lambda =0 \\\\ & \Rightarrow 5 \lambda-3 =0 \\\\ & \Rightarrow \lambda =\frac{3}{5} \end{aligned}

Putting the value of $\lambda$ in Eq. (i), we get

\begin{aligned} & (2 x-y+z-3)+\frac{3}{5}(4 x-3 y+5 z+9)=0 \\\\ & \Rightarrow 11 x-7 y+10 z+6=0 \\\\ & \therefore a=11, b=-7, c=10 \text{ and } d=6 \\\\ & \therefore a+b+c=11-7+10=14 \end{aligned}

Q188

One vertex of a rectangular parallelopiped is at the origin

\mathrm{O}

and the lengths of its edges along

x, y

and

z

axes are

3,4

and

5

units respectively. Let

\mathrm{P}

be the vertex

(3,4,5)

. Then the shortest distance between the diagonal OP and an edge parallel to

\mathrm{z}

axis, not passing through

\mathrm{O}

or

\mathrm{P}

is :

A

\dfrac{12}{\sqrt{5}}

B

12 \sqrt{5}

C

\dfrac{12}{5}

D

\dfrac{12}{5 \sqrt{5}}

Correct Answer

Option C

Solution

Equation of $O P$ is

\begin{aligned} &\frac{x-0}{3-0}=\frac{y-0}{4-0} =\frac{z-0}{5-0} \\\\ &\Rightarrow \frac{x}{3} =\frac{y}{4}=\frac{z}{5} \end{aligned}

Equation of edge parallel to $Z$ -axis is

\frac{x-3}{0}=\frac{y-0}{0}=\frac{z-5}{1}

$\therefore$ Shortest distance

=\frac{\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right|}{\sqrt{\begin{array}{r} \left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2 +\left(c_1 a_2-c_2 a_1\right)^2 \end{array}}}

Here,

\begin{aligned} & x_1=0, y_1=0, z_1=0 \\\\ & x_2=3, y_2=0, z_2=5 \\\\ & a_1=3, b_1=4, c_1=5 \\\\ & a_2=0, b_2=0, c_2=1 \end{aligned}

\therefore\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right|=\left|\begin{array}{ccc} 3 & 0 & 5 \\ 3 & 4 & 5 \\ 0 & 0 & 1 \end{array}\right|=4(3)=12

$\begin{aligned} & \text{ and } \sqrt{\left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2} \\\\ & =\sqrt{0+(4-0)^2+(0-3)^2} \\\\ & =\sqrt{16+9}=5 \\\\ & \therefore \text{ Required shortest distance }=\dfrac{12}{5} \\\\ & \end{aligned}$

Q189

A plane P contains the line of intersection of the plane

\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6

and

\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5

. If

\mathrm{P}

passes through the point

(0,2,-2)

, then the square of distance of the point

(12,12,18)

from the plane

\mathrm{P}

is :

A 310

B 620

C 1240

D 155

Correct Answer

Option B

Solution

Given plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ and

\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5

Equation of plane passing through both plane

\begin{aligned} & \mathrm{P}_1 \rightarrow(x \hat{i}+y \hat{j}+2 \hat{k})(\hat{i}+\hat{j}+\hat{k})=6 \\\\ & \mathrm{P}_1=x+y+z=6 \\\\ & \mathrm{P}_2 \rightarrow(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5 \\\\ & \mathrm{P}_2 \rightarrow=2 x+3 y+4 z=-5 \end{aligned}

\begin{aligned} & \mathrm{P}_1+\lambda \mathrm{P}_2=0 \\\\ & \Rightarrow(x+y+z-6)+\lambda(2 x+3 y+4 z+5)=0 \end{aligned}

passes through $(0,2,-2)$

\begin{aligned} & \Rightarrow(0+2-2-6)+\lambda(2 \times 0+3 \times 2+4 \times(-2)+5)=0 \\\\ & \Rightarrow \lambda=2 \end{aligned}

Equation of plane

5 x+7 y+9 z+4=0

$\therefore$ Distance of the point $(12,12,18)$ from the plane

\begin{aligned} d & =\frac{5 \times 12+7 \times 12+9 \times 18+4}{\sqrt{5^2+7^2+9^2}} \\\\ & =\frac{60+84+162+4}{\sqrt{155}}=\frac{310}{\sqrt{155}} \\\\ \therefore d^2 & =\frac{310 \times 310}{155}=620 \end{aligned}

Q190

Let the line

\mathrm{L}

pass through the point

(0,1,2)

, intersect the line

\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}

and be parallel to the plane

2 x+y-3 z=4

. Then the distance of the point

\mathrm{P}(1,-9,2)

from the line

\mathrm{L}

is :

A 9

B

\sqrt{74}

C

\sqrt{69}

D

\sqrt{54}

Correct Answer

Option B

Solution

Given line, $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{1}=\lambda$ .........(i) $\therefore$ Any point on the line $A(2 \lambda+1,3 \lambda+2, \lambda+3)$ Since, line $L$ pass through the point $B(0,1,2)$ and intersects line (i) $\therefore$ Direction ratio of $A B$ are

\begin{array}{ll} & 0-2 \lambda-1,1-3 \lambda-2,2-\lambda-3 \\\\ &\text{ i.e., } -2 \lambda-1,-3 \lambda-1,-\lambda-1 \end{array}

Line $L$ parallel to the plane $2 x+y-3 z=4$

\begin{aligned} & \therefore 2(-2 \lambda-1)+1(-3 \lambda-1)-3(-\lambda-1)=0 \\\\ & \Rightarrow-4 \lambda-2-3 \lambda-1+3 \lambda+3=0 \\\\ & \Rightarrow -4 \lambda=0 \\\\ & \Rightarrow \lambda=0 \\\\ & \therefore \text{ Direction ratio of } A B=-1,-1,-1 \\\\ & \text{ i.e., } 1,1,1 \end{aligned}

$\therefore$ Equation of line $L$ is $\dfrac{x-0}{1}=\dfrac{y-1}{1}=\dfrac{z-2}{1}=\mu$

\Rightarrow x=\mu, y=\mu+1, z=\mu+2

$\therefore$ Direction ratios of $P Q=\mu-1, \mu+10, \mu$ Since, $P Q \perp L$

\begin{aligned} & \therefore 1(\mu-1)+1(\mu+10)+\mu(1)=0 \\\\ & \Rightarrow \mu-1+\mu+10+\mu=0 \\\\ & \Rightarrow 3 \mu+9=0 \\\\ & \Rightarrow \mu=-3 \\\\ & \therefore Q \text{ is }(-3,-2,-1) \end{aligned}

$\therefore$ Distance of the point $P(1,-9,2)$ from the line $L$ is

\begin{array}{r} = \sqrt{(-3-1)^2+(-2+9)^2+(-1-2)^2} \\\\ = \sqrt{16+49+9}=\sqrt{74} \end{array}