1
Question
The distance of line 3y−2z−1=0=3x−z+4 from the point (2,−1,6) is
The distance of line 3y−2z−1=0=3x−z+4 from the point (2,−1,6) is
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Solution
The correct option is A 2√6
Direction ratio's of given line
∣∣
∣
∣∣^i^j^k03−230−1∣∣
∣
∣∣=^i(−3)−^j(6)+^k(−9)
=−3^i−6^j−9^k
Let z=0⇒y=13 and x=−43
∴ Line in Cartesian form is
x+43−3=y−13−6=z−9
Let point of shortest distance be P(λ) i.e.
P(−λ−43,−2λ+13,−3λ) and Q(2,−1,6)
For shortest distance −−→PQ⋅(^i+2^j+3^k)=0
((103+λ)^i+(2λ−43)^j+(6+3λ)^k)⋅(^i+2^j+3^k)=0
⇒ λ=−43
∴P≡(0,3,4)
∴|PQ|=2√6
Direction ratio's of given line
∣∣ ∣ ∣∣^i^j^k03−230−1∣∣ ∣ ∣∣=^i(−3)−^j(6)+^k(−9)
=−3^i−6^j−9^k
Let z=0⇒y=13 and x=−43
∴ Line in Cartesian form is
x+43−3=y−13−6=z−9
Let point of shortest distance be P(λ) i.e.
P(−λ−43,−2λ+13,−3λ) and Q(2,−1,6)
For shortest distance −−→PQ⋅(^i+2^j+3^k)=0
((103+λ)^i+(2λ−43)^j+(6+3λ)^k)⋅(^i+2^j+3^k)=0
⇒ λ=−43
∴P≡(0,3,4)
∴|PQ|=2√6
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