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 · 25 Jan · Shift 1 · Q31
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 · 25 Jan · Shift 1 · Q31

3D Geometry question

2023 · 25 Jan · Shift 1 · Q31

JEE MainMathematics3D GeometryMCQ+4 / −1
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 :
  1. A
    3
  2. B
    14\sqrt{14}14​
  3. C
    6\sqrt66​
  4. D
    232\sqrt323​
View written solutionFree

Correct answer: B

  1. Write the line in vector form

The required line passes through the point A(−3,2,3)A(-3,2,3)A(−3,2,3) and is parallel to a line with direction ratios 3,3,−1.3,3,-1.3,3,−1. So a direction vector of the line is d⃗=(3,3,−1).\vec d=(3,3,-1).d=(3,3,−1).

  1. Form the vector from a point on the line to the given point

Given point: P(4,6,−2).P(4,6,-2).P(4,6,−2).

Hence, AP→=P−A=(4−(−3), 6−2, −2−3)=(7,4,−5).\overrightarrow{AP}=P-A=(4-(-3),\,6-2,\,-2-3)=(7,4,-5).AP=P−A=(4−(−3),6−2,−2−3)=(7,4,−5).

  1. Use the formula for distance of a point from a line in 3D

Distance of point PPP from the line through AAA with direction vector d⃗\vec dd is Distance=∣AP→×d⃗∣∣d⃗∣.\text{Distance}=\frac{\left|\overrightarrow{AP}\times \vec d\right|}{|\vec d|}.Distance=∣d∣​AP×d​​.

So first compute

\begin{vmatrix} \hat i & \hat j & \hat k \\ 7 & 4 & -5 \\ 3 & 3 & -1 \end{vmatrix}.$$ Expanding: $$=\hat i\,(4\cdot(-1)-(-5)\cdot 3)-\hat j\,(7\cdot(-1)-(-5)\cdot 3)+\hat k\,(7\cdot 3-4\cdot 3).$$ $$=\hat i(-4+15)-\hat j(-7+15)+\hat k(21-12)$$ $$=11\hat i-8\hat j+9\hat k.$$ Thus, $$\left|\overrightarrow{AP}\times \vec d\right|=\sqrt{11^2+(-8)^2+9^2}= \sqrt{121+64+81}= \sqrt{266}.$$ Also, $$|\vec d|=\sqrt{3^2+3^2+(-1)^2}= \sqrt{9+9+1}= \sqrt{19}.$$ Therefore, $$\text{Distance}=\frac{\sqrt{266}}{\sqrt{19}}= \sqrt{\frac{266}{19}}= \sqrt{14}.$$ 4. **Match with options** $$\sqrt{14}$$ corresponds to **Option B**. 5. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** So they agree.
PreviousNext

More from 3D Geometry

  • 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
  • If the shortest distance between the line joining the points (1, 2, 3) and (2, 3, 4), and the line 2x−1​=−1y+1​=0z−2​ is α, then 28 α2 is equal to ​.2023 · Numerical
  • Let the co-ordinates of one vertex of ΔABC be A(0,2,α) and the other two vertices lie on the line 5x+α​=2y−1​=3z+4​. For α∈Z, if the area of ΔABC…2023 · Numerical
  • The shortest distance between the lines 2x−1​=−7y+8​=5z−4​ and 2x−1​=1y−2​=−3z−6​ is :2023 · MCQ
  • Let a line L pass through the point P(2,3,1) and be parallel to the line x+3y−2z−2=0=x−y+2z. If the distance of L from the point (5,3,8) is α, then 3α2 is equal to :2023 · Numerical
  • Let the shortest distance between the lines L:−2x−5​=0y−λ​=1z+λ​,λ≥0 and L1​:x+1=y−1=4−z be 26​. If (α,β,γ) lies on L, then which of the following is…2023 · MCQ