Find the angle between the plares $r . (2 i - 3 j + 4 k)$ and $r . (- i + j) = 4$

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Solution

Given r.(2i-3j+4k)=k ...(1)
r.(-i+j)=4 ...(2)
angle between plane (1)and (2)is equal to angle
between to direction ratios
direction ratios of Plane 1 =2i-3j+4k = $n_{1}$
directions ratios of plane 2 =-i+j = $n_{2}$
$c o s θ = ∣ ∣ \frac{n_{1} - n_{2}}{| n_{1} | | n_{2} |} ∣ ∣ = \frac{(2 i - 3 j + 4 k) (- i + j)}{\sqrt{4 + 9 + 16} \sqrt{1 + 1}}$
$c o s θ = ∣ ∣ ∣ \frac{- 2 - 3}{\sqrt{29} \sqrt{2}} ∣ ∣ ∣ = \frac{5}{\sqrt{58}}$
angle $θ = c o s^{- 1} (\frac{5}{\sqrt{58}})$ is angle between given plane

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