Show that a vector coplanar with - - - - 2 k- and - - 2 - - k- and perpendicular to - - - - k- is - 1 - 2 - - k- -

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Standard XII

Mathematics

Condition for Coplanarity of Four Points

Show that a v...

Question

Show that a vector coplanar with $^i +^j + 2^k$ and $^i + 2^j +^k$ and perpendicular to $^i +^j +^k$ is $\pm \frac{1}{\sqrt{2}} (^j -^k)$ .

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Solution

Let
$\begin{matrix} \to a = ˆ i + ˆ j + 2 ˆ k \to b = ˆ i + 2 ˆ j + ˆ k \to n = ˆ i + ˆ j + ˆ k \to c = x \to a + y \to b \to c \cdot \to n = 0 = x (\to a \cdot \to n) + y (\to b \cdot \to n) \Rightarrow x (4) + y (4) = 0 y = - x \to c = x (\to a - \to b) = x (- ˆ j + ˆ k) ∣ ∣ \to c ∣ ∣ = 1 \sqrt{x^{2} + x^{2}} = 1 \sqrt{2 x^{2}} = 1 2 x^{2} = 1 x^{2} = \frac{1}{2} x = \pm \frac{1}{\sqrt{2}} \to c = \pm \frac{1}{\sqrt{2}} (- ˆ j + ˆ k) \Rightarrow \frac{1}{\sqrt{2}} (ˆ j - ˆ k), \frac{1}{\sqrt{2}} (ˆ k - ˆ j) \end{matrix}$

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