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If vector equation of the line ๐‘ฅ โˆ’ 2/2 = 2โข๐‘ฆ โˆ’ 5/โˆ’3 = ๐‘ง + 1 is ฬ…๐‘Ÿ = (2โขฬ‚๐‘– + 5/2โขฬ‚๐‘— โˆ’ฬ‚๐‘˜) + ๐œ†โข(2โขฬ‚๐‘– โˆ’ 3/2โขฬ‚๐‘— + ๐‘โขฬ‚๐‘˜) then p is equal to ______.

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Question

If vector equation of the line `(x - 2)/2 = (2y - 5)/-3 = z + 1  "is"  barr = (2hati + 5/2 hatj - hatk) + lambda (2hati - 3/2 hatj + phatk)` then p is equal to ______.

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Solution

If vector equation of the line `(x - 2)/2 = (2y - 5)/-3 = z + 1  "is"  barr = (2hati + 5/2 hatj - hatk) + lambda (2hati - 3/2 hatj + phatk)` then p is equal to 0.

Explanation:

The given line is

`(x - 2)/2 = (2y - 5)/-3 = z + 1`,

= `(x - 2)/2 = (y - 5/2)/(-3/2) = (z + 1)/0`

This shows that the given line passes through the point `(2, 5/2, -1)` and has direction ratios `(2, -3/2, 0)`. Thus, given line passes through the point having position vector `bara = 2hati + 5/2 hatj - hatk` and is parallel to the vector `barb = (2hati - 3/2 hatj - 0hatk)`.

So, its vector equation is `barr = (2hati + 5/2 hatj - hatk) + lambda (2hati - 3/2 hatj - 0hatk)`.

Hence, p = 0

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