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3D Geometry question

2019 · 8 Apr · Shift 2 · Q31
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3D Geometry question

2019 · 8 Apr · Shift 2 · Q31

JEE MainMathematics3D GeometryMCQ+4 / −1
If a point R(4, y, z) lies on the line segment joining the points P(2, –3, 4) and Q(8, 0, 10), then the distance of R from the origin is :
  1. A
    2142 \sqrt {14}214​
  2. B
    53\sqrt {53}53​
  3. C
    2212 \sqrt {21}221​
  4. D
    6
View written solutionFree

Correct answer: A

  1. Write the parametric form of the line through P(2,−3,4)P(2,-3,4)P(2,−3,4) and Q(8,0,10)Q(8,0,10)Q(8,0,10).

A point on the line segment PQPQPQ can be written as

R=P+t(Q−P),0≤t≤1.R=P+t(Q-P), \qquad 0\le t\le 1.R=P+t(Q−P),0≤t≤1.

Now,

Q−P=(8−2,  0−(−3),  10−4)=(6,3,6).Q-P=(8-2,\;0-(-3),\;10-4)=(6,3,6).Q−P=(8−2,0−(−3),10−4)=(6,3,6).

So,

R=(2,−3,4)+t(6,3,6).R=(2,-3,4)+t(6,3,6).R=(2,−3,4)+t(6,3,6).

Hence the coordinates of RRR are

R=(2+6t, −3+3t, 4+6t).R=(2+6t,\,-3+3t,\,4+6t).R=(2+6t,−3+3t,4+6t).
  1. Use the given first coordinate of RRR.

We are told that R=(4,y,z)R=(4,y,z)R=(4,y,z). Therefore,

2+6t=4.2+6t=4.2+6t=4.

So,

6t=2  ⟹  t=13.6t=2 \implies t=\frac{1}{3}.6t=2⟹t=31​.
  1. Find yyy and zzz.

Substitute t=13t=\frac13t=31​:

y=−3+3(13)=−3+1=−2,y=-3+3\left(\frac13\right)=-3+1=-2,y=−3+3(31​)=−3+1=−2, z=4+6(13)=4+2=6.z=4+6\left(\frac13\right)=4+2=6.z=4+6(31​)=4+2=6.

Thus,

R=(4,−2,6).R=(4,-2,6).R=(4,−2,6).
  1. Compute the distance of RRR from the origin O(0,0,0)O(0,0,0)O(0,0,0).

Distance formula:

OR=42+(−2)2+62.OR=\sqrt{4^2+(-2)^2+6^2}.OR=42+(−2)2+62​.

So,

OR=16+4+36=56=214.OR=\sqrt{16+4+36}= \sqrt{56}=2\sqrt{14}.OR=16+4+36​=56​=214​.
  1. Match with the options.
2142\sqrt{14}214​

corresponds to Option A.

Therefore, the correct answer is A.

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