Find the equation of the plane passing through the line of intersection of the planes r-3-6 - 0 and r-3-4k- - 0 whose perpendicular distance from origin is unity-

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Byju's Answer

Standard XII

Mathematics

Family of Planes Passing through the Intersection of Two Planes

Find the equa...

Question

Find the equation of the plane passing through the line of intersection of the planes $\to r . (^i + 3^j) - 6 = 0$ and $\to r . (3^i -^j - 4^k) = 0$ whose perpendicular distance from origin is unity.

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Solution

The equation of the plane passing through the line intersection of the plane $\to r . (\to i + 3 \to j) - 6 = 0$ and $\to r . (3 \to i - \to j - 4 \to k) = 0$ is $\to r [(\to i + 3 \to j) + λ (3 \to i - \to j - 4 \to k)] = 6 + λ (0)$ [Using $\to r . [_{1} + λ_{2}] = d_{1} + λ_{2}$ ]
i.e., $\to r . [(1 + 3 λ) \to i + (3 - λ) \to j - 4 λ \to k] - 6 = 0$ ---- (1)
Now distnace of (1) from (0, 0) is $\frac{∣ ∣ ∣ (o \to i + 0 \to j + 0 \to k [((1 + 3 λ) \to i + (3 - λ) \to j - 4 λ \to k] - 6)]) ∣ ∣ ∣}{\sqrt{(1 + 3 λ)^{2} + (3 - λ)^{2} + (- 4 λ)^{2}}} = 1$
${(\frac{6}{\sqrt{(1 + 3 λ)^{2} + (3 - λ)^{2} + (- 4 λ)^{2}}})}^{2} = 1 \Rightarrow \frac{36}{26 λ^{2} + 10} = 1 \Rightarrow λ = \pm 1$
Substituting the value of $λ$ in (1), we get $\to r . [4 \to i + 2 \to j - 4 \to k] - 6 = 0$
i.e., $\to r . [2 \to i + \to j - 2 \to k] = 3$
and $\to r . [- 2 \to i + 4 \to j + 4 \to k] - 6 = 0$ i.e., $\to r . [\to i - 2 \to j - 2 \to k] + 3 = 0$

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Find the equation of the plane passing through the line of intersection of the planes →r.(^i+3^j)−6=0 and →r.(3^i−^j−4^k)=0 whose perpendicular distance from origin is unity.

Find the equation of the plane passing through the line of intersection of the planes $\to r . (^i + 3^j) - 6 = 0$ and $\to r . (3^i -^j - 4^k) = 0$ whose perpendicular distance from origin is unity.