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CBSE Class 12-science Answered

Find the coordinates of the point, where the line (x-2)/3 = (y+1)/4 = (z-2)/2 intersects the plane x - y + z - 5 = 0. Also, find the angle between the line and the plane.
Asked by EFFY OOMMEN JOHN | 08 Feb, 2017, 12:30: PM
answered-by-expert Expert Answer
begin mathsize 16px style Let space the space coordinates space of space the space point space on space the space line space fraction numerator straight x minus 2 over denominator 3 end fraction equals fraction numerator straight y plus 1 over denominator 4 end fraction equals fraction numerator straight z minus 2 over denominator 2 end fraction equals straight r
So comma space straight x equals 3 straight r plus 2 comma space space space straight y space equals space 4 straight r minus 1 comma space space space straight z equals 2 straight r plus 2
If space it space lies space on space the space plane space straight x minus straight y plus straight z minus 5 equals 0 comma space
then space 3 straight r plus 2 minus space left parenthesis 4 straight r minus 1 right parenthesis plus 2 straight r plus 2 minus 5 equals 0 rightwards double arrow straight r equals 0
Substituting space this space in space straight x equals 3 straight r plus 2 comma space space space straight y space equals space 4 straight r minus 1 comma space space space straight z equals 2 straight r plus 2 comma space we space can space obtain space the space coordinates.
So comma space straight x equals 2 comma space straight y equals negative 1 comma space straight z equals 2
Hence comma space the space coordinates space of space the space point space of space intersection space of space the space line space and space space space are space left parenthesis 2 comma space minus 1 comma space 2 right parenthesis.
The space given space line space is space parallel space to space the space vector space straight a with rightwards arrow on top equals 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top space space and space the space given space plane space is space normal space to space the space vector space straight n with rightwards arrow on top equals straight i with hat on top minus straight j with hat on top plus straight k with hat on top
Recall space that space sinθ equals fraction numerator straight a with rightwards arrow on top. straight n with rightwards arrow on top over denominator vertical line straight a with rightwards arrow on top open vertical bar straight n with rightwards arrow on top close vertical bar end fraction equals fraction numerator open parentheses 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top space close parentheses. open parentheses straight i with hat on top minus straight j with hat on top plus straight k with hat on top close parentheses over denominator vertical line 3 straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top open vertical bar straight i with hat on top minus straight j with hat on top plus straight k with hat on top close vertical bar end fraction equals fraction numerator 3 minus 4 plus 2 over denominator square root of 9 plus 16 plus 4 end root square root of 1 plus 1 plus 1 end root end fraction equals fraction numerator 1 over denominator square root of 87 end fraction
rightwards double arrow sinθ equals fraction numerator 1 over denominator square root of 87 end fraction
Hence comma space straight theta equals sin to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 87 end fraction close parentheses. end style
Answered by Rebecca Fernandes | 08 Feb, 2017, 08:16: PM
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