1
Question
Find the vector equation of the plane which bisects the line segment joining the points A(5,7,2) and B(−1,−3,4)
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Solution
−−→AB=(−^i−3^j+4^k)−(5^i+7^j+2^k)
=−6^i−10^j+2^k(=→n)
Let P be the mid-point of AB. Then position vector or P=(5^i+7^j+2^k)(−^i−3^j+4^k)2
=2^i+2^j+3^k(=→d)
The equation plane passing through P and normal to AB. Hence, its vector equation is given by,
(→r−→d).→n=0
→r.→n=→d.→n
→r.(−6^i−10^j+2^k)=(2^i+2^j+3^k).(−6^i−10^j+2^k)
=(2)(−6)+2(−10)+(3)(2)=−12−20+6=−26
⇒→r.(−6^i−10^j+2^k)=−26
⇒→r.(3^i+5^j−^k)=13
=−6^i−10^j+2^k(=→n)
Let P be the mid-point of AB.
Then position vector or P=(5^i+7^j+2^k)(−^i−3^j+4^k)2
=2^i+2^j+3^k(=→d)
The equation plane passing through P and normal to AB. Hence, its vector equation is given by,
(→r−→d).→n=0
→r.→n=→d.→n
→r.(−6^i−10^j+2^k)=(2^i+2^j+3^k).(−6^i−10^j+2^k)
=(2)(−6)+2(−10)+(3)(2)=−12−20+6=−26
⇒→r.(−6^i−10^j+2^k)=−26
⇒→r.(3^i+5^j−^k)=13
=2^i+2^j+3^k(=→d)
The equation plane passing through P and normal to AB. Hence, its vector equation is given by,
(→r−→d).→n=0
→r.→n=→d.→n
→r.(−6^i−10^j+2^k)=(2^i+2^j+3^k).(−6^i−10^j+2^k)
=(2)(−6)+2(−10)+(3)(2)=−12−20+6=−26
⇒→r.(−6^i−10^j+2^k)=−26
⇒→r.(3^i+5^j−^k)=13
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