First slide

Straight lines in 3D

Question

Consider the lines x = y = z and the line 2x + y + z –1 = 0 = 3x + y + 2z – 2 is

Moderate

A

The shortest distance between the two lines is $\frac{1}{\sqrt{2}}$
B

The shortest distance between the two lines is $\sqrt{2}$
C

Plane containing 2nd line parallel to 1st line is
D

The shortest distance between the two lines $\frac{\sqrt{3}}{2}$

Solution

Any plane through the second line is 2x + y + z –1 + k (3x + y + 2z –2) = 0
If this is parallel to the first line, then $(2 + 3 k) + (1 + k) + (1 + 2 k) = 0$ $\Rightarrow k = - \frac{2}{3}$
$\Rightarrow P l a n e i s 2 x + y + z - 1 - \frac{2}{3} (3 x + y + 2 z - 2) = 0$ or $y - z + 1 = 0.$
The required SD must be distance of this plane from any point on the line $x = y = z$ say (1, 1, 1)
$\Rightarrow SD = \frac{| 1 - 1 + 1 |}{\sqrt{0^{2} + 1^{2} + {(- 1)}^{2}}} = \frac{1}{\sqrt{2}}$

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