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Question: Distance of the point \(\left( x _ { 1 } , y _ { 1 } , z _ { 1 } \right)\) from the line \(\frac {...

Distance of the point (x1,y1,z1)\left( x _ { 1 } , y _ { 1 } , z _ { 1 } \right) from the line x−x2l=y−y2m=z−z2n\frac { x - x _ { 2 } } { l } = \frac { y - y _ { 2 } } { m } = \frac { z - z _ { 2 } } { n }, where ll m and n are the direction cosines of line is

A

(x1−x2)2+(y1−y2)2+(z1−z2)2−[(1x1−x2)+m(y1−y2)+n(z1−z2)]2\sqrt { \left( x _ { 1 } - x _ { 2 } \right) ^ { 2 } + \left( y _ { 1 } - y _ { 2 } \right) ^ { 2 } + \left( z _ { 1 } - z _ { 2 } \right) ^ { 2 } - \left[ \left( 1 x _ { 1 } - x _ { 2 } \right) + m \left( y _ { 1 } - y _ { 2 } \right) + n \left( z _ { 1 } - z _ { 2 } \right) \right] ^ { 2 } }

B

(x2−x1)2+(y2−y1)2+(z2−z1)2\sqrt { \left( x _ { 2 } - x _ { 1 } \right) ^ { 2 } + \left( y _ { 2 } - y _ { 1 } \right) ^ { 2 } + \left( z _ { 2 } - z _ { 1 } \right) ^ { 2 } }

C

(x2−x1)l+(y2−y1)m+(z2−z1)n\sqrt { \left( x _ { 2 } - x _ { 1 } \right) l + \left( y _ { 2 } - y _ { 1 } \right) m + \left( z _ { 2 } - z _ { 1 } \right) n }

D

None of these

Answer

(x1−x2)2+(y1−y2)2+(z1−z2)2−[(1x1−x2)+m(y1−y2)+n(z1−z2)]2\sqrt { \left( x _ { 1 } - x _ { 2 } \right) ^ { 2 } + \left( y _ { 1 } - y _ { 2 } \right) ^ { 2 } + \left( z _ { 1 } - z _ { 2 } \right) ^ { 2 } - \left[ \left( 1 x _ { 1 } - x _ { 2 } \right) + m \left( y _ { 1 } - y _ { 2 } \right) + n \left( z _ { 1 } - z _ { 2 } \right) \right] ^ { 2 } }

Explanation

Solution

Let r1=(x2−x1)i+(y2−y1)j+(z2−z1)k\mathbf { r } _ { 1 } = \left( x _ { 2 } - x _ { 1 } \right) \mathbf { i } + \left( y _ { 2 } - y _ { 1 } \right) \mathbf { j } + \left( z _ { 2 } - z _ { 1 } \right) \mathbf { k } r2=li+mj+nk\mathbf { r } _ { 2 } = l \mathbf { i } + m \mathbf { j } + n \mathbf { k }

∴\therefore cos⁡θ=r2⋅r1∣r1∣∣r2∣\cos \theta = \frac { \mathbf { r } _ { 2 } \cdot \mathbf { r } _ { 1 } } { \left| \mathbf { r } _ { 1 } \right| \left| \mathbf { r } _ { 2 } \right| }

Also, d=∣r1∣sin⁡θd = \left| \mathbf { r } _ { 1 } \right| \sin \theta ,

⇒

⇒ d2=∣r1∣2(1−r1⋅r2∣r1∣2∣r2∣2)d ^ { 2 } = \left| \mathbf { r } _ { 1 } \right| ^ { 2 } \left( 1 - \frac { \mathbf { r } _ { 1 } \cdot \mathbf { r } _ { 2 } } { \left| \mathbf { r } _ { 1 } \right| ^ { 2 } \left| \mathbf { r } _ { 2 } \right| ^ { 2 } } \right)

⇒ , {where ∣r2∣=1\left| \mathbf { r } _ { 2 } \right| = 1}

⇒

Therefore, distance of the point (x1,y1,z1x _ { 1 } , y _ { 1 } , z _ { 1 }) from the line is

d =.