Single Choice

The shortest distance between the lines $$\dfrac {x}{2} = \dfrac {y}{2} = \dfrac {z}{1}$$ and $$\dfrac {x + 2}{-1} = \dfrac {y - 4}{8} = \dfrac {z - 5}{4}$$ lies in the interval:

A$$[1, 2)$$
B$$(3, 4]$$
C$$[0, 1)$$
D$$(2, 3]$$
Correct Answer

Solution

Writing the equation of Line in Vector Form

$$r_1=\lambda (2i+2j+k)$$
$$r_2=(-2i+4j+5k) + \lambda (-i+8j+4k)$$
$$a_2-a_1= (-2i+4j+5k)$$
$$b_1 \times b_2=(2i+2j+k) \times (-i+8j+4k)$$
$$b_1 \times b_2= -9j+18k$$

Now,

$$(a_2-a_1) . (b_1 \times b_2) = (-2i+4j+5k) . (-9j+18k) = 54$$

$$|b_1 \times b_2|= 9\sqrt5$$
Shortest Distance $$ d= |\dfrac{{(a_2-a_1)} . {(b_1 \times b_2)}}{|(b_1 \times b_2)|}| $$

Hence, $$d= \dfrac{54}{9\sqrt5}=\dfrac{6}{\sqrt5}$$ which lies between $$(2,3]$$

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