3
Question
Assertion :L1:x+13=y+21=z+12, L2:x−21=y+22=z−33
The distance of the point (1,1,1) from the plane passing through the point (−1,−2,−1) and whose normal is perpendicular to both the lines L1 and L2 is 135√3. Reason: The unit vector perpendicular to both the lines L1 and L2 is −→i−7→j+5→k5√3.
Assertion :L1:x+13=y+21=z+12, L2:x−21=y+22=z−33
The distance of the point (1,1,1) from the plane passing through the point (−1,−2,−1) and whose normal is perpendicular to both the lines L1 and L2 is 135√3. Reason: The unit vector perpendicular to both the lines L1 and L2 is −→i−7→j+5→k5√3.
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Solution
The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
Normal vector of the plane is given by,
→n=∣∣
∣
∣∣^i^j^k312123∣∣
∣
∣∣=−^i−7^j+5^k
Hence, equation of plane is,
(→r−(−^i−2^j−1^k))⋅→n=0
⇒−(x+1)−7(y+2)+5(z+1)=0
⇒−x−7y+5z−10=0....(p)
Hence, perpendicular distance from (1,1,1) is
=∣−1−7+5−10√12+72+52∣=135√3
Also normal vector perpendicular to both L1 and L2 is
=−^i−7^j+5^k5√3
Since shortest distance between two line is along unit vector perpendicular to both line. Hence, Reason is correct explanation of Assertion.
Normal vector of the plane is given by,
→n=∣∣ ∣ ∣∣^i^j^k312123∣∣ ∣ ∣∣=−^i−7^j+5^k
Hence, equation of plane is,
(→r−(−^i−2^j−1^k))⋅→n=0
⇒−(x+1)−7(y+2)+5(z+1)=0
⇒−x−7y+5z−10=0....(p)
Hence, perpendicular distance from (1,1,1) is
=∣−1−7+5−10√12+72+52∣=135√3
Also normal vector perpendicular to both L1 and L2 is
=−^i−7^j+5^k5√3
Since shortest distance between two line is along unit vector perpendicular to both line. Hence, Reason is correct explanation of Assertion.
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