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Family of Planes Passing through the Intersection of Two Planes

Determine the...

Question

Determine the equation of plane through the point $a = (i + 2 j - k)$ and prependicular to the line of intersection of planes $r . (3 i - j + k) = 1$ and $r . (i + 4 j - 2 k) = 2.$

r . (- 2 i + 7 j + 13 k) = 1.

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r . (- 2 i + 7 j + 13 k) = - 1.

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r . (6 i - 5 j + 11 k) = - 1.

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r . (- 6 i + 5 j - 11 k) = 1.

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Solution

The correct option is B $r . (- 2 i + 7 j + 13 k) = - 1.$
$r . n_{1} = q_{1}, r n_{2} = q_{2}$ are two planes.
The line of intersection of these plance will be perpendicular to both
$n_{1}$ and $n_{2}$ and hence parallel to $n_{1} \times n_{2} .$
$n_{1} \times n_{2} = N = ∣ ∣ ∣ ∣ \begin{matrix} i & j & k 3 & - 1 & 1 1 & 4 & - 2 \end{matrix} ∣ ∣ ∣ ∣ = - 2 i + 7 i + 13 k$
Since the given plane is perpendicular to $N$ and hence its normal is $N$ and it passes through a.
Hence its equation is $(r - a) . N = 0$ $\to r . N = a . N$
$\Rightarrow r . (- 2 i + 7 j + 13 k) = 1 (- 2) + 2 (7) - 1 (13) .$ $\Rightarrow r . (- 2 i + 7 j + 13 k) = - 1.$

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r . (3 i - j + k) = 1

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l_{1} : r (t) = (i - 6 j + 2 k) + t (i + 2 j + k)

l_{2} : R (u) = (4 j + k) + u (2 i + j + 2 k)

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