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Question
Find the vector equation of the line which passes through the origin and the point (5, –2, 3).
Solution
Let `bar"b"` be the position vector of the point B(5, –2, 3).
Then `bar"b" =5hat"i" - 2hat"j" + 3hat"k"`
Origin has position vector `bar"0" = 0hat"i" + 0hat"j" + 0hat"k"`.
The vector equation the line passing through `"A"(bara) and "b"(barb) "is" bar"r" = bar"a" + lambda(bar"b" - bar"a")` where λ is a scalar.
∴ the vector equation of the required line is
`bar"r" = bar"0" + lambda(bar"b" - bar"0") = lambda(5hat"i" - 2hat"j" + 3hat"k")`.
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