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Question

Distance (in units) of a point with position vector ^i+2^j+^k from the line given by L:r=2^i+^j+3^k+λ(^i+^j+^k) is:

A
143
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B
73
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C
143
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D
73
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Solution

The correct option is A 143
We know
d=|(pa)×c||c|
Given: L:r=2^i+^j+3^k+λ(^i+^j+^k),p=^i+2^j+^k
Comparing with r=a+λc, we have (a)=2^i+^j+3^k and c=^i+^j+^k
Here, (pa)=^i+^j2^k
Now, (pa)×c=∣ ∣ ∣^i^j^k112111∣ ∣ ∣=3^i^j2^k
|c|=3
d=|(pa)×c||c|=|3^i^j2^k|3
Thus, d=143 units

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