Misc 14

Transcript

Misc 14 If the points (1, 1 , p) and (–3 , 0, 1) be equidistant from the plane 𝑟﷯. (3 𝑖﷯ + 4 𝑗﷯ − 12 𝑘﷯) + 13 = 0, then find the value of p. The distance of a point with position vector 𝑎﷯ from the plane 𝑟﷯. 𝑛﷯ = d is 𝒂﷯. 𝒏﷯ − 𝒅﷮ 𝒏﷯﷯﷯﷯ Given, the points are The equation of plane is 𝑟﷯. (3 𝑖﷯ + 4 𝑗﷯ − 12 𝑘﷯) + 13 = 0 𝑟﷯.(3 𝑖﷯ + 4 𝑗﷯ − 12 𝑘﷯) = −13 – 𝑟﷯.(3 𝑖﷯ + 4 𝑗﷯ − 12 𝑘﷯) = 13 𝒓﷯.(–3 𝒊﷯ – 4 𝒋﷯ + 12 𝒌﷯) = 13 Comparing with 𝑟﷯. 𝑛﷯ = d, 𝑛﷯ = −3 𝑖﷯ − 4 𝑗﷯ + 12 𝑘﷯ & d = 13 Magnitude of 𝑛﷯ = ﷮ −3﷯﷮2﷯+ −4﷯﷮2﷯+ 12﷮2﷯﷯ 𝑛﷯﷯ = ﷮9+16+144﷯ = ﷮169﷯ = 13 Since the plane is equidistance from both the points, 𝟏𝟐𝒑 − 𝟐𝟎﷮𝟏𝟑﷯﷯ = 𝟖﷮𝟏𝟑﷯ 12𝑝−20﷯ = 8 (12p – 20) = ± 8 Solving So, p = 𝟕﷮𝟑﷯ & p = 1 (Answer does not match at end. If mistake, please comment)