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Question
Find the vector equation of the line which passes through the origin and intersect the line x – 1 = y – 2 = z – 3 at right angle.
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Solution
The given line is `(x - 1)/(1) = (y - 2)/(1) = (z - 3)/(1) = lambda` ...(Say)
∴ coordinates of any point on the line are x = `lambda + 1, y = lambda + 2, z = lambda + 3`
∴ position vector of any point on the line is `(lambda + 1)hat"i" + (lambda + 2)hat"j" + (lambda + 3)hat"k"` ...(1)
If `bar"b"` is parallel to the given line whose direction ratios are 1, 1, 1 then `bar"b" = hat"i" + hat"j" + hat"k"`.
Let the required line passing through O meet the given line at M.
∴ position vector of M
= `bar"m" = (lambda + 1)hat"i" + (lambda + 2)hat"j" + (lambda + 3)hat"k"` ...[By (1)]
The required line is perpendicular to given line
∴ `bar"OM".bar"b"` = 0
∴ `[(lambda + 1)hat"i" + (lambda + 2)hat"j" + (lambda + 3)hat"k"].(hat"i" + hat"j" + hat"k")` = 0
∴ `(lambda + 1) xx 1 + (lambda + 2) xx 1 + (lambda + 3) xx 1` = 0
∴ `3lambda + 6` = 0
∴ λ = – 2
∴ `bar"m" = (-2 + 1)hat"i" + (-2 + 2)hat"j" + (-2 + 3)hat"k" = -hat"i" + hat"k"`
The vector equation of the line passing through `"A"(bara) and "B"(barb) "is" bar"r" = bar"a" + lambda(bar"b" - bar"a"), lambda` is a scalar.
∴ the vector equation of the line passing through `"O"(bar"0") and "M"(bar"m") "is" bar"r" = bar"0" + lambda(bar"m" - bar"0") = lambda(-hat"i" + hat"k")` where λ is a scalar.
Hence, vector equation of the required line is `bar"r" = lambda(-hat"i" + hat"k")`.
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