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Question 147 of 162

Q.Find the vector equation of a plane which is at a distance of 7 units from the origin having 3,−4,53, -4, 5 as direction ratios of a normal to it.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2023Subjective· 2mImportance★★★★★
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Scales the given normal direction to a unit vector and writes the plane's vector equation r⃗⋅n^=d\vec r\cdot\hat n=d with d=7d=7.

  1. Normal direction ratios are 3,−4,53,-4,5, so n⃗=3i^−4j^+5k^\vec n=3\hat i-4\hat j+5\hat k, with ∣n⃗∣=32+(−4)2+52=9+16+25=50=52|\vec n|=\sqrt{3^2+(-4)^2+5^2}=\sqrt{9+16+25}=\sqrt{50}=5\sqrt2.
  2. Unit normal: n^=3i^−4j^+5k^52\hat n=\dfrac{3\hat i-4\hat j+5\hat k}{5\sqrt2}. …

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