Question

Let P be a plane passing through the points (1, 0, 1), (1, –2, 1) and (0, 1, –2). Let a vector `vec"a" = αhat"i" + βhat"j" + γhat"k"` be such that `veca` is parallel to the plane P, perpendicular to `(hat"i"+2hat"j"+3hat"k")`and `vec"a".(hat"i" + hat"j" + 2hat"j")` = 2, then (α – β + γ)² equals ______.

Options

• 80

• 81

• 82

• 83

Answer 1

Let P be a plane passing through the points (1, 0, 1), (1, –2, 1) and (0, 1, –2). Let a vector `vec"a" = αhat"i" + βhat"j" + γhat"k"` be such that `vec"a"` is parallel to the plane P, perpendicular to `(hat"i"+2hat"j"+3hat"k")`and `vec"a".(hat"i" + hat"j" + 2hat"j")` = 2, then (α – β + γ)² equals 81.

Explanation:

Equation of plane P is `|(x - 1, "y" - 0, "z" - 1),(1 - 1, 2, 1 - 1),(1 - 0, 0 - 1, 1 + 2)|` = 0

or, `|(x - 1, "y" - 0, "z" - 1),(0, 2, 0),(1, -1, 3)|` = 0

⇒ 6(x –1) + (z – 1)(–2) = 0

⇒ 3x – z – 2 = 0

Normal vector to the plane P is `vec"n" = 3hat"i" - hat"k"`

Now, `vec"a" = αhat"i" + βhat"j" + γhat"k"` is perpendicular to `vec"n"`

⇒ `vec"a".vec"n"` = 0

⇒ 3α – γ = 0.   ...(i)

Also, `vec"a"`. is perpendicular to `vec"b" = hat"i" + 2hat"j" + 3hat"k"`

⇒ `vec"a".vec"b"` = 0

⇒ α + 2β + 3γ = 0  ...(ii)

And `vec"a".(hat"i" + hat"j" + 2hat"k")` = 2

⇒ α + β + 2γ = 2  ...(iii)

By solving (i), (ii) and (iii)

⇒ α = 1, β = –5, γ = 3

⇒  (α – β + γ)² = 81