A line l passing through the origin is perpendicular to the lines
l1:(3+t)^i+(−1+2t)^j+(4+2t)^k,−∞<t<∞l2:(3+2s)^i+(3+2s)^j+(2+s)^k,−∞<s<∞
Then, the coordinate(s) of the point(s) on l2 at a distance of √17 from the point of intersection of l and l1 is (are)
A line l passing through the origin is perpendicular to the lines
l1:(3+t)^i+(−1+2t)^j+(4+2t)^k,−∞<t<∞l2:(3+2s)^i+(3+2s)^j+(2+s)^k,−∞<s<∞
Then, the coordinate(s) of the point(s) on l2 at a distance of √17 from the point of intersection of l and l1 is (are)
The correct options are
B −1,−1,0
D (79,79,89)
Equation of straight line is l:x−x1a=y−y1b=z−z1c
which is perpendicular to l1 and l2
So, its DR's are cross-product of l1 and l2
Now, to find a point on l2 whose distance is given, assume a point and find its distance to obtian point
Let l:x−0a=y−0b=z−0c
which is perpendicular to
l1:(3^i−^j+4^k)+t(^i+2^j+2^k)l2:(3^i+3^j+2^k)+s(2^i+2^j+^k)∴ DR's of l is∣∣
∣
∣∣^i^j^k122221∣∣
∣
∣∣=−2^i+3^j−2^kl:x−2=y3=z−2=k1,k2Now, A(−2k1,3k1,−2k1) and B(−2k2,3k2,−2k2).
Since, A lies on l1.
∴(−2k1)^i+(3k1)^j−(2k1)^k=(3+t)^i+(−1+2t)^j+(4+2t)^k⇒3+t=−2k1,−1+2t=3k1,4+2t=−2k1∴ k1=−1⇒A(2,−3,2)
Let any point on l2(3+2s,3+2s,2+s)
√(2−3−2s)2+(−3−3−2s)2+(2−2−s)2=√17⇒9s2+28s+37=17⇒9s2+28s+20=0⇒9s2+18s+10s+20=0⇒(9s+10)(s+2)=0∴s=−2,−109.Hence,(−1,−1,0) and (79,79,89) are required points.
B
−1,−1,0
D
(79,79,89)
Equation of straight line is l:x−x1a=y−y1b=z−z1c
which is perpendicular to l1 and l2
So, its DR's are cross-product of l1 and l2
Now, to find a point on l2 whose distance is given, assume a point and find its distance to obtian point
Let l:x−0a=y−0b=z−0c
which is perpendicular to
l1:(3^i−^j+4^k)+t(^i+2^j+2^k)l2:(3^i+3^j+2^k)+s(2^i+2^j+^k)∴ DR's of l is∣∣
∣
∣∣^i^j^k122221∣∣
∣
∣∣=−2^i+3^j−2^kl:x−2=y3=z−2=k1,k2Now, A(−2k1,3k1,−2k1) and B(−2k2,3k2,−2k2).
Since, A lies on l1.
∴(−2k1)^i+(3k1)^j−(2k1)^k=(3+t)^i+(−1+2t)^j+(4+2t)^k⇒3+t=−2k1,−1+2t=3k1,4+2t=−2k1∴ k1=−1⇒A(2,−3,2)
Let any point on l2(3+2s,3+2s,2+s)
√(2−3−2s)2+(−3−3−2s)2+(2−2−s)2=√17⇒9s2+28s+37=17⇒9s2+28s+20=0⇒9s2+18s+10s+20=0⇒(9s+10)(s+2)=0∴s=−2,−109.Hence,(−1,−1,0) and (79,79,89) are required points.
