A straight line r-a-b- meets the plane r-n-0 in R then P-V of R is given by

The correct option is A a̅ a̅.n̅b.n.b̅ r̅.n̅ = 0⇒ (a̅+λb̅)·n̅ = 0⇒λ = a⃗·n⃗b⃗·n⃗ Hence point of intersection is, r⃗ = a⃗+λb⃗ = a⃗ a⃗·n⃗b⃗·n⃗b⃗ ...

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Standard XII

Mathematics

Section Formula

A straight li...

Question

A straight line $¯ r = ¯ a + λ ¯ b$ meets the plane $¯ r . ¯ n = 0$ in R then P.V of R is given by

¯ a - \frac{¯ a . ¯ n}{b . n} . ¯ b

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2 ¯ a - \frac{¯ a . ¯ n}{¯ b . ¯ n} . ¯ b

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¯ a + \frac{¯ a . ¯ n}{¯ b . ¯ n} . ¯ b

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None of these

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Solution

The correct option is A $¯ a - \frac{¯ a . ¯ n}{b . n} . ¯ b$
$¯ r . ¯ n = 0 \Rightarrow (¯ a + λ ¯ b) \cdot ¯ n = 0 \Rightarrow λ = - \frac{\to a \cdot \to n}{\to b \cdot \to n}$
Hence point of intersection is, $\to r = \to a + λ \to b = \to a - \frac{\to a \cdot \to n}{\to b \cdot \to n} \to b$

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A straight line ¯r=¯a+λ¯b meets the plane ¯r.¯n=0 in R then P.V of R is given by

The correct option is A ¯a−¯a.¯nb.n.¯b¯r.¯n=0⇒(¯a+λ¯b)⋅¯n=0⇒λ=−→a⋅→n→b⋅→nHence point of intersection is, →r=→a+λ→b=→a−→a⋅→n→b⋅→n→b

A straight line $¯ r = ¯ a + λ ¯ b$ meets the plane $¯ r . ¯ n = 0$ in R then P.V of R is given by