A line L passes through a point with position vector a and is perpendicular to the plane containing the two intersecting lines r-b-p and r-c-q- where - If L intersects a plane r-n-d at the point R- then the position vector of the point R is

Equation of the line L is nbsp;r=a+t(p×q) Since L nbsp;intersects plane nbsp;r·n=d, ∴(a+t(p×q))·n=d ⇒a·n+t[p q n]=d ⇒ t=(d a·n)[p q n] Thus, the posit ...

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Byju's Answer

Standard XII

Mathematics

Condition for a Line to Lie on a Plane

A line L pass...

Question

A line $L$ passes through a point with position vector $\to a$ and is perpendicular to the plane containing the two intersecting lines $\to r = \to b + λ \to p$ and $\to r = \to c + μ \to q,$ where $λ, μ \in R .$ If $L$ intersects a plane $\to r \cdot \to n = d$ at the point $R,$ then the position vector of the point $R$ is

(d - \to a \cdot \to n) (\to p \times \to q)

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\to a + \frac{(d - \to a \cdot \to n)}{[\to p \to q \to n]} (\to p \times \to q)

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\to a + [\to p \to q \to n] \to b + [\to p \to q \to a] \to c

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\to a + \frac{(d + \to a \cdot \to n)}{[\to p \to q \to n]} (\to p \times \to q)

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Solution

The correct option is B $\to a + \frac{(d - \to a \cdot \to n)}{[\to p \to q \to n]} (\to p \times \to q)$
Equation of the line $L$ is $\to r = \to a + t (\to p \times \to q)$
Since $L$ intersects plane $\to r \cdot \to n = d,$
$∴ (\to a + t (\to p \times \to q)) \cdot \to n = d$
$\Rightarrow \to a \cdot \to n + t [\to p \to q \to n] = d$
$\Rightarrow t = \frac{(d - \to a \cdot \to n)}{[\to p \to q \to n]}$
Thus, the position vector of $R$ is
$\to r = \to a + \frac{(d - \to a \cdot \to n)}{[\to p \to q \to n]} (\to p \times \to q)$

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