Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

3D Geometry question

2023 · 12 Apr · Shift 1 · Q26
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /3D Geometry
  5. /2023 · 12 Apr · Shift 1 · Q26

3D Geometry question

2023 · 12 Apr · Shift 1 · Q26

JEE MainMathematics3D GeometryMCQ+4 / −1
Let the lines l1:x+53=y+41=z−α−2l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}l1​:3x+5​=1y+4​=−2z−α​ and l2:3x+2y+z−2=0=x−3y+2z−13l_{2}: 3 x+2 y+z-2=0=x-3 y+2 z-13l2​:3x+2y+z−2=0=x−3y+2z−13 be coplanar. If the point P(a,b,c)\mathrm{P}(a, b, c)P(a,b,c) on l1l_{1}l1​ is nearest to the point Q(−4,−3,2)\mathrm{Q}(-4,-3,2)Q(−4,−3,2), then ∣a∣+∣b∣+∣c∣|a|+|b|+|c|∣a∣+∣b∣+∣c∣ is equal to
  1. A
    12
  2. B
    14
  3. C
    10
  4. D
    8
View written solutionFree

Correct answer: C

  1. Write the lines in usable form

For l1l_1l1​, x+53=y+41=z−α−2=t\frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}=t3x+5​=1y+4​=−2z−α​=t So its parametric form is x=−5+3t,y=−4+t,z=α−2t.x=-5+3t,\quad y=-4+t,\quad z=\alpha-2t.x=−5+3t,y=−4+t,z=α−2t. A point on l1l_1l1​ is A(−5,−4,α),A(-5,-4,\alpha),A(−5,−4,α), and its direction vector is d⃗1=(3,1,−2).\vec d_1=(3,1,-2).d1​=(3,1,−2).

For l2l_2l2​, it is the intersection of planes 3x+2y+z−2=0andx−3y+2z−13=0.3x+2y+z-2=0 \quad \text{and} \quad x-3y+2z-13=0.3x+2y+z−2=0andx−3y+2z−13=0. Its direction vector is the cross product of the plane normals: n⃗1=(3,2,1),n⃗2=(1,−3,2).\vec n_1=(3,2,1),\quad \vec n_2=(1,-3,2).n1​=(3,2,1),n2​=(1,−3,2). Hence

\begin{vmatrix} \hat i & \hat j & \hat k\\ 3&2&1\\ 1&-3&2 \end{vmatrix} =(7,-5,-11).$$ 2. **Find a point on $l_2$** Let $z=0$. Then from the plane equations, $$3x+2y=2,\qquad x-3y=13.$$ From the second, $$x=13+3y.$$ Substitute into the first: $$3(13+3y)+2y=2$$ $$39+11y=2$$ $$y=-\frac{37}{11}.$$ Then $$x=13+3\left(-\frac{37}{11}\right)=\frac{32}{11}.$$ So a point on $l_2$ is $$B\left(\frac{32}{11},-\frac{37}{11},0\right).$$ 3. **Use coplanarity condition to find $\alpha$** Two lines are coplanar if $$[\overrightarrow{AB},\vec d_1,\vec d_2]=0,$$ where $$\overrightarrow{AB}=B-A=\left(\frac{32}{11}+5,-\frac{37}{11}+4,-\alpha\right)=\left(\frac{87}{11},\frac{7}{11},-\alpha\right).$$ Thus, $$\det\begin{pmatrix} \frac{87}{11} & \frac{7}{11} & -\alpha\\ 3&1&-2\\ 7&-5&-11 \end{pmatrix}=0.$$ Expand along the first row: $$\frac{87}{11} \begin{vmatrix} 1&-2\\ -5&-11 \end{vmatrix} -\frac{7}{11} \begin{vmatrix} 3&-2\\ 7&-11 \end{vmatrix} +(-\alpha) \begin{vmatrix} 3&1\\ 7&-5 \end{vmatrix}=0.$$ Now, $$\begin{vmatrix}1&-2\\-5&-11\end{vmatrix}=-11-10=-21,$$ $$\begin{vmatrix}3&-2\\7&-11\end{vmatrix}=-33+14=-19,$$ $$\begin{vmatrix}3&1\\7&-5\end{vmatrix}=-15-7=-22.$$ So, $$\frac{87}{11}(-21)-\frac{7}{11}(-19)+(-\alpha)(-22)=0$$ $$-166+\frac{133}{11}+22\alpha=0$$ $$-166+\frac{133}{11}+22\alpha=0.$$ Since $$-166=-\frac{1826}{11},$$ we get $$\frac{-1826+133}{11}+22\alpha=0$$ $$-\frac{1693}{11}+22\alpha=0$$ $$22\alpha=\frac{1693}{11}$$ $$\alpha=\frac{1693}{242}=7.$$ So $l_1$ is $$x=-5+3t,\quad y=-4+t,\quad z=7-2t.$$ 4. **Find the point on $l_1$ nearest to $Q(-4,-3,2)$** Let $$P(t)=(-5+3t,-4+t,7-2t).$$ Then $$\overrightarrow{QP}=P-Q=(-5+3t+4,-4+t+3,7-2t-2)=(3t-1,t-1,5-2t).$$ For nearest point, $\overrightarrow{QP}$ must be perpendicular to direction vector $(3,1,-2)$: $$ (3t-1,t-1,5-2t)\cdot(3,1,-2)=0.$$ Compute: $$3(3t-1)+(t-1)-2(5-2t)=0$$ $$9t-3+t-1-10+4t=0$$ $$14t-14=0$$ $$t=1.$$ Hence $$P=(-5+3,-4+1,7-2)=(-2,-3,5).$$ So $$a=-2,\quad b=-3,\quad c=5.$$ Therefore, $$|a|+|b|+|c|=2+3+5=10.$$ 5. **Compare with options** The value is $$\boxed{10}.$$ So the correct option is **C**.
PreviousNext

More from 3D Geometry

  • The line, that is coplanar to the line −3x+3​=1y−1​=5z−5​, is :2023 · MCQ
  • Let S be the set of all values of λ, for which the shortest distance between the lines 0x−λ​=4y−3​=1z+6​ and 3x+λ​=−4y​=0z−6​ is 13. Then 8​λ∈S∑​λ​…2023 · MCQ
  • The shortest distance between the lines 3x−2​=2y+1​=2z−6​ and 3x−6​=21−y​=0z+8​ is equal to ​2023 · Numerical
  • If the shortest between the lines 2x+6​​=3y−6​​=4z−6​​ and 3x−λ​=4y−26​​=5z+26​​ is 6, then the square of sum of…2023 · Numerical
  • The distance of the point P(4, 6, − 2) from the line passing through the point (− 3, 2, 3) and parallel to a line with direction ratios 3, 3, − 1 is equal to :2023 · MCQ
  • Consider the lines L1​ and L2​ given by L1​:2x−1​=1y−3​=2z−2​L2​:1x−2​=2y−2​=3z−3​. A line L3​ having direction ratios 1, − 1, − 2,…2023 · MCQ
  • The foot of perpendicular of the point (2, 0, 5) on the line 2x+1​=5y−1​=−1z+1​ is (α,β,γ). Then, which of the following is NOT correct?2023 · MCQ
  • The shortest distance between the lines x+1=2y=−12z and x=y+2=6z−6 is :2023 · MCQ