3D Geometry — Page 24 | JEE Mathematics

Q231

If the image of the point

(4,4,3)

in the line

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

is

(\alpha, \beta, \gamma)

, then

\alpha+\beta+\gamma

is equal to

A 12

B 9

C 7

D 8

Correct Answer

Option B

Solution

\begin{aligned} &\begin{aligned} & \overrightarrow{\mathrm{PQ}} \perp(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & \Rightarrow 2(2 \lambda-3)+1(\lambda-2)+3(3 \lambda-2)=0 \\ & \Rightarrow 14 \lambda-14=0, \lambda=1 \\ & \text{ So, } \mathrm{Q}(3,3,4) \end{aligned}\\ &\text{ Let image in } R(\alpha, \beta, \gamma)\\ &\begin{aligned} & \frac{\alpha+4}{2}=3, \frac{\beta+4}{2}=3, \frac{\gamma+3}{2}=4 \\ & (\alpha, \beta, \gamma)=(2,2,5) \\ & \Rightarrow \alpha+\beta+\gamma=9 \end{aligned} \end{aligned}

Q232

Let A be the point of intersection of the lines

\mathrm{L}_1: \dfrac{x-7}{1}=\dfrac{y-5}{0}=\dfrac{z-3}{-1}

and

\mathrm{L}_2: \dfrac{x-1}{3}=\dfrac{y+3}{4}=\dfrac{z+7}{5}

. Let B and C be the points on the lines

\mathrm{L}_1

and

\mathrm{L}_2

respectively such that

A B=A C=\sqrt{15}

. Then the square of the area of the triangle

A B C

is :

A 63

B 57

C 60

D 54

Correct Answer

Option D

Solution

$L_1: \dfrac{x-7}{1}=\dfrac{y-5}{0}=\dfrac{z-3}{-1} ; L_2: \dfrac{x-1}{3}=\dfrac{y+3}{4}=\dfrac{z+7}{5}$

\begin{aligned} & \cos \theta=\left|\frac{3+0-5}{\sqrt{2} \times \sqrt{50}}\right| \\ & =\frac{2}{10}=\frac{1}{5} \\ & \therefore \sin \theta=\frac{2 \sqrt{6}}{5} \\ & \text{ Area }=\frac{1}{2} a b \sin \theta \\ & =\frac{1}{2} \times \sqrt{15} \times \sqrt{15} \times \frac{2 \sqrt{6}}{5} \\ & =3 \sqrt{6} \\ & (\text{ Area })^2=9 \times 6=54 \end{aligned}

Q233

Let the shortest distance between the lines

\dfrac{x-3}{3}=\dfrac{y-\alpha}{-1}=\dfrac{z-3}{1}

and

\dfrac{x+3}{-3}=\dfrac{y+7}{2}=\dfrac{z-\beta}{4}

be

3 \sqrt{30}

. Then the positive value of

5 \alpha+\beta

is

A 42

B 40

C 48

D 46

Correct Answer

Option D

Solution

\begin{aligned} & L_1: \frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1} \\ & L_2: \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4} \end{aligned}

\begin{aligned} & a_1:(3, \alpha, 3) \quad a^2=(-3,-7, \beta) \\ & \vec{b}_1=3 \hat{i}-\hat{j}+\hat{k} \quad \vec{b}_2=-3 \hat{i}+2 \hat{j}+4 \hat{k} \end{aligned}

$d=\dfrac{\left|\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\vec{b}_1 \times \vec{b}_2\right|}=3 \sqrt{30}$ $=\dfrac{\left|\begin{array}{ccc}6 & \alpha+7 & 3-\beta \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{array}\right|}{\sqrt{6^2+15^2+3^2}}=3 \sqrt{30}$

\begin{aligned} \Rightarrow & |-15 \alpha-3 \beta-132|=270 \\ & |5 \alpha+\beta+44|=90 \\ \Rightarrow & 5 \alpha+\beta=90-44=46 \end{aligned}

Q234

Let the values of

\lambda

for which the shortest distance between the lines

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

and

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

is

\dfrac{1}{\sqrt{6}}

be

\lambda_1

and

\lambda_2

. Then the radius of the circle passing through the points

(0, 0), (\lambda_1, \lambda_2)

and

(\lambda_2, \lambda_1)

is

A

3

B

\dfrac{5\sqrt{2}}{3}

C

\dfrac{\sqrt{2}}{3}

D

4

Correct Answer

Option B

Solution

\begin{aligned} &\begin{aligned} & \overrightarrow{\mathrm{p}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \overrightarrow{\mathrm{q}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\ & \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & \mathrm{~A} \equiv(1,2,3) \mathrm{B} \equiv(\lambda, 4,5) \\ & \text{ Shortest Distance }=\left|\frac{\overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\right| \\ & \frac{1}{\sqrt{6}}=\left|\frac{(\lambda-1) \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})}{\sqrt{6}}\right| \\ & \Rightarrow|-\lambda+1+4-2|=1 \Rightarrow|\lambda-3|=1 \\ & \Rightarrow \lambda=3 \pm 1=4,2 \end{aligned}\\ &\text{ Radius of circle passing through points }\\ &\begin{aligned} & (0,0),(4,2) \&(2,4) \\ & =\frac{a b c}{4 \Delta}=\frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times \frac{1}{2}\left|\begin{array}{lll} 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 \end{array}\right|}=\frac{20 \times 2 \sqrt{2}}{2 \times 12} \\ & =\frac{5 \sqrt{2}}{3} \end{aligned} \end{aligned}

Q235

Let the vertices Q and R of the triangle PQR lie on the line

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

and the coordinates of the point

P

be

(0,2,3)

. If the area of the triangle

P Q R

is

\dfrac{m}{n}

then :

A

2 \mathrm{~m}-5 \sqrt{21} \mathrm{n}=0

B

\mathrm{m}-5 \sqrt{21} \mathrm{n}=0

C

5 \mathrm{~m}-21 \sqrt{2} \mathrm{n}=0

D

5 \mathrm{~m}-2 \sqrt{21} \mathrm{n}=0

Correct Answer

Option A

Solution

\begin{aligned} & H:(5 \lambda-3,2 \lambda+1,3 \lambda-4) \\ & <\mathrm{DR} \text{ of } \mathrm{PH}> \\ & <5 \lambda-3,2 \lambda-1,3 \lambda-7> \\ & \overrightarrow{P H} \cdot \overrightarrow{Q R}=0 \\ & \Rightarrow(5 \lambda-3) 5+(2 \lambda-1) 2+(3 \lambda-7) 3=0 \\ & \Rightarrow 25 \lambda-15+4 \lambda-2+9 \lambda-21=0 \\ & \Rightarrow 38 \lambda=38 \\ & \Rightarrow \lambda=1 \\ & \therefore \quad H(2,3,-1) \\ & P H=\sqrt{4+1+16}=\sqrt{21} \\ & \therefore \quad \text{ Area }=\frac{1}{2} \times P H \text{ QR } \\ & \quad=\frac{1}{2} \times \sqrt{21} \times 5=\frac{5 \sqrt{21}}{2}=\frac{m}{n} \\ & 2 m-5 \sqrt{21} n=0 \end{aligned}

Q236

Let

A B C D

be a tetrahedron such that the edges

A B, A C

and

A D

are mutually perpendicular. Let the areas of the triangles

\mathrm{ABC}, \mathrm{ACD}

and ADB be 5,6 and 7 square units respectively. Then the area (in square units) of the

\triangle B C D

is equal to :

A

\sqrt{110}

B 12

C

\sqrt{340}

D

7 \sqrt{3}

Correct Answer

Option A

Solution

\begin{aligned} & \operatorname{ar}(\triangle A B C)=5 \\ & \frac{1}{2} \times b c=5 \\ & \Rightarrow \quad b c=10 \\ & \operatorname{ar}(\triangle A C D)=6 \\ & \frac{1}{2} \times c d=6 \\ & \Rightarrow \quad c d=12 \\ & \operatorname{ar}(\triangle A B D)=7 \\ & b d=14 \end{aligned}

\begin{aligned} & \operatorname{area}(\triangle B C D)=\frac{1}{2}|\overrightarrow{B C} \times \overrightarrow{B D}| \\ & \overrightarrow{B C}=<-b, c, 0> \\ & \overrightarrow{B D}=<-b, 0, d> \\ & |\overrightarrow{B C} \times \overrightarrow{B D}|=\sqrt{c^2 d^2+b^2 d^2+b^2 c^2} \\ & =\sqrt{12^2+14^2+10^2} \\ & =\sqrt{440} \\ & \operatorname{ar}(\triangle B C D)=\sqrt{110} \end{aligned}

Q237

If the equation of the line passing through the point

\left( 0, -\dfrac{1}{2}, 0 \right)

and perpendicular to the lines

\vec{r} = \lambda \left( \hat{i} + a\hat{j} + b\hat{k} \right)

and

\vec{r} = \left( \hat{i} - \hat{j} - 6\hat{k} \right) + \mu \left( -b \hat{i} + a\hat{j} + 5\hat{k} \right)

is

\dfrac{x-1}{-2} = \dfrac{y+4}{d} = \dfrac{z-c}{-4}

, then

a+b+c+d

is equal to :

A 13

B 14

C 12

D 10

Correct Answer

Option B

Solution

The line we need to find should be orthogonal to two given lines, meaning it will be parallel to the cross product of their direction vectors: $(\hat{i} + a \hat{j} + b \hat{k}) \times (-b \hat{i} + a \hat{j} + 5 \hat{k})$ Calculating the cross product results in the vector: $\hat{i}(5a - ab) - \hat{j}(b^2 + 5) + \hat{k}(a + ab)$ This shows that the direction ratios of the required line are proportional to: $\alpha(5a - ab), - (b^2 + 5), (a + ab)$ Since the line is also expressed as $\dfrac{x-1}{-2} = \dfrac{y+4}{d} = \dfrac{z-c}{-4}$ , the direction ratios based on this equation are: $\alpha(-2), d, -4$ Thus, we have the equation: $\dfrac{5a - ab}{-2} = \dfrac{- (b^2 + 5)}{d} = \dfrac{a + ab}{-4}$ The point $\left(0, -\dfrac{1}{2}, 0 \right)$ lies on the line, which implies: $\dfrac{0-1}{-2} = \dfrac{-\dfrac{1}{2} + 4}{d} = \dfrac{0-c}{-4}$ Simplifying these gives $d = 7$ and $c = 2$ .

Substitute into the main equation: $\dfrac{5a - ab}{-2} = \dfrac{a + ab}{-4}$ $\dfrac{-b^2 - 5}{7} = \dfrac{a + ab}{-4}$ Solving these yields: $\begin{array}{c|c} -20a + 4ab = -2a - 2ab & 4b^2 + 20 = 70 + 7ab \\ 18a = 6ab & 36 + 20 = 70 + 21a \\ b = 3 & 56 = 28a \Rightarrow a = 2 \end{array}$ Substituting these into the variables, we find: $a + b + c + d = 2 + 3 + 2 + 7 = 14$

Q238

Let a line passing through the point

(4,1,0)

intersect the line

\mathrm{L}_1: \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}

at the point

A(\alpha, \beta, \gamma)

and the line

\mathrm{L}_2: x-6=y=-z+4

at the point

B(a, b, c)

. Then

\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c\end{array}\right|

is equal to

A 16

B 6

C 8

D 12

Correct Answer

Option C

Solution

\begin{aligned} & A=(\alpha, \beta, 4)=(2 \lambda+1,3 \lambda+2,4 \lambda+3) \\ & \text{ and } B=(a, b, c)=(\mu+6, \mu,-\mu+4) \\ & \because \frac{\mu+2}{2 \lambda-3}=\frac{\mu-1}{3 \lambda+1}=\frac{-\mu+4}{4 \lambda+3} \\ & \therefore \quad \lambda \mu+8 \lambda+4 \mu=1 \quad \ldots(1)\\ & \text{ and } 7 \lambda \mu-16 \lambda+4 \mu=7 \quad \ldots(2) \end{aligned}

from equation (1) and (2)

\lambda=-1 \text{ and } \mu=3

Or $\lambda=-\dfrac{1}{3}$ and $\mu=1$ By taking, $\lambda=-1$ and $\mu=3$ , we get

\begin{aligned} & \therefore \quad A(\alpha, \beta, 4)=(-1,-1,-1) \\ & \text{ and } B(a, b, c)=(9,3,1) \end{aligned}

\therefore\left|\begin{array}{ccc} 1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1 \end{array}\right|=8

Q239

The two lines

x=ay+b,z=cy+d

and

x = a'y + b',z = c'y + d'

will be perpendicular, if and only if :

A

aa' + cc' + 1 = 0

B

aa' + bb'cc' + 1 = 0

C

aa' + bb'cc' = 0

D

\left( {a + a'} \right)\left( {b + b'} \right) + \left( {c + c'} \right) = 0

Correct Answer

Option A

Solution

{{x - b} \over a} = {y \over 1} = {{z - d} \over c};

{{x - b'} \over {a'}}

= {y \over 1} = {{z - d'} \over c'}

For perpenedicularity of lines

aa' + 1 + cc' = 0

Q240

Consider the lines L1: x - 1 = y - 2 = z and L2: x - 2 = y = z - 1. Let the feet of the perpendiculars from the point P(5, 1, -3) on the lines L1 and L2 be Q and R respectively. If the area of the triangle PQR is A, then 4A2 is equal to :

A 151

B 147

C 139

D 143

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{L}_1: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-0}{2} \\ & \text{ Let } \mathrm{Q}(\lambda+1, \lambda+2, \lambda) \\ & \overrightarrow{\mathrm{PQ}}=(\lambda-4, \lambda-1, \lambda+3) \\ & \overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{~m}}=0 \end{aligned}

\begin{aligned} &\begin{aligned} & \Rightarrow \lambda-4+\lambda+1, \lambda+3=0 \\ & \Rightarrow 3 \lambda=0 \\ & \Rightarrow \lambda=0 \\ & \Rightarrow \mathrm{Q}(1,2,0) \\ & \mathrm{L}_2: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{z}-1}{2} \end{aligned}\\ &\text{ Let } \mathrm{R}(\mu+2, \mu, \mu+1) \overrightarrow{\mathrm{PR}}=(\mu-3, \mu-1, \mu+4)\\ &\overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{n}}=0\\ &\mu-3+\mu-1+\mu+4=0\\ &\neq \mu=0\\ &\mathrm{R}(2,0,1) \end{aligned}

\begin{aligned} & \text{ Area of } \triangle \mathrm{PQR}(\mathrm{~A})=\frac{1}{2}|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PR}}| \\ & \mathrm{A}=\frac{1}{2}|(-4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times(-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}})| \\ & \mathrm{A}=\frac{1}{2}|7(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})| \\ & \left|\begin{array}{lrr} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -4 & 1 & 3 \\ -3 & -1 & 4 \end{array}\right| \\ & =7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & 4 \mathrm{~A}^2=49 \times 3=147 \end{aligned}