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Equation of a Plane : Normal Form

If a⃗ = -i ...

Question

If $\to a = - i + j + k$ and $\to b = 2 i + k$ , then the vector $\to c$ satisfying the conditions.
(i) that it is coplanar with $\to a$ and $\to b$
(ii) that its projection on $\to b$ is $0$ .

- 3 i + 5 j + 6 k

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- 3 i - 5 j + 6 k

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- 6 i + 5 k

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- i + 2 j + 2 k

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Solution

The correct option is A $- 3 i + 5 j + 6 k$
Given: $\to a = - i + j + k$ and $\to b = 2 i + k$
Find: $\to c$ such that i) it is coplanar with $\to a$ and $\to b$
and ii) it's projection on $\to b$ is 0
Sol: let $\to c = x i + y j + z k$
since $\to c$ is coplanar with $\to a$ and $\to b$
$⟹ \to c, (\to a x \to b) = 0$
$⟹ \to c [(- i + j + k) x (2 i + k)] = 0$
$⟹ \to c [i + 3 j - 2 k] = 0$
$⟹ (x i + y j + z k) (i + 3 j - 2 k) = 0$
$⟹ x + 3 y - 2 z = 0 . . . . . . . . . . (i)$
projection of $\to c$ on $\to b = \frac{\to c . \to b}{\to b} = 0$
$⟹ c . b = 0 ⟹ 2 x + z = 0 . . . . (i i)$
From (i) and (ii) we get that
$x + 3 y - 2 (- 2 x) = 0$
$⟹ 5 x + 3 y = 0 ⟹ y = \frac{- 5 x}{3}$ and $z = 2 x$
Therefore $\to c = x i + (\frac{- 5 x}{3}) j + (2 x) k$
$= \frac{x}{3} [3 i - 5 j - 6 k] = \frac{- x}{3} [- 3 i + 5 j + 6 k]$
Hence, correct answer is $\to c = [- 3 i + 5 j + 6 k]$

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If →a=−i+j+k and →b=2i+k, then the vector →c satisfying the conditions.(i) that it is coplanar with →a and →b(ii) that its projection on →b is 0.

If $\to a = - i + j + k$ and $\to b = 2 i + k$ , then the vector $\to c$ satisfying the conditions.
(i) that it is coplanar with $\to a$ and $\to b$
(ii) that its projection on $\to b$ is $0$ .