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Question
The shortest distance between the lines
x−33=y−8−1=z−31 and
x+3−3=y+72=z−64 is:
The shortest distance between the lines
x−33=y−8−1=z−31 and
x+3−3=y+72=z−64 is:
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Solution
The correct option is D 3√30
−−→AB=−3^i−7^j+6^k−(3^i+8^j+3^k)=−6^i−15^j+3^k
→p=3^i−^j+^k
→q=−3^i+2^j+4^k
→p×→q=∣∣
∣
∣∣^i^j^k3−11−324∣∣
∣
∣∣=−6^i−15^j+3^k
Shortest distance =|−−→AB⋅(→p×→q)||→p×→q|=|36+225+9|√36+225+9=3√30.
−−→AB=−3^i−7^j+6^k−(3^i+8^j+3^k)=−6^i−15^j+3^k
→p=3^i−^j+^k
→q=−3^i+2^j+4^k
→p×→q=∣∣ ∣ ∣∣^i^j^k3−11−324∣∣ ∣ ∣∣=−6^i−15^j+3^k
Shortest distance =|−−→AB⋅(→p×→q)||→p×→q|=|36+225+9|√36+225+9=3√30.
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