Q. 15

# Find the eq

Answer :

The equation of any plane through the line of intersection of the planes and is given by - .

So, the equation of any plane through the line of intersection of the given planes is . . . …(1)

Since this plane is parallel to x-axis.

So, the normal vector of the plane (1) will be perpendicular to x-axis.

Direction ratios of Normal (a1, b1, c1)≡ [(1 - 2λ), (1 - 3λ), (1 +)]

Direction ratios of x–axis (a2, b2, c2)≡ (1,0,0)

Since the two lines are perpendicular,

a1a2 + b1b2 + c1c2 = 0

(1 - 2λ) × 1 + (1 - 3λ) × 0 + (1 + λ) × 0 = 0

(1 - 2λ) = 0

λ = 1/2

Putting the value of λ in (1), we get -   Hence, the equation of the required plane is Rate this question :

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