Find the distance of a point (2, 5, −3) from the plane $straight r with rightwards arrow on top. open parentheses 6 straight i with hat on top minus 3 straight j with hat on top plus 2 straight k with hat on top close parentheses space equals 4.$

Solution

Consider the vector equation of the plane.
$straight r with rightwards arrow on top. space open parentheses 6 straight i with hat on top minus 3 straight j with hat on top space plus space 2 straight k with hat on top close parentheses space equals space 4 space space space space space rightwards double arrow open parentheses straight x straight i with hat on top space plus space straight y straight j with hat on top space plus space straight z straight k with hat on top close parentheses. space open parentheses 6 straight i with hat on top minus 3 straight j with hat on top space plus space 2 straight k with hat on top close parentheses space equals space 4 space space space space space rightwards double arrow 6 straight x minus 37 plus 2 straight z space equals 4 space space space space space rightwards double arrow 6 straight x minus 3 straight y plus 2 straight z minus 4 space equals space 0 space space space space space space space$
Thus the Cartesian equation of the plane is
6x-3y+2z-4 = 0
Let d be the distance between the point (2, 5, -3)
$Thus space straight d space equals space open vertical bar fraction numerator ax subscript 1 plus by subscript 1 plus cz subscript 1 plus straight d over denominator square root of straight a squared plus straight b squared plus straight c squared end root end fraction close vertical bar space space space rightwards double arrow straight d space equals space open vertical bar fraction numerator 6 cross times 2 minus 3 cross times 5 plus 2 cross times left parenthesis negative 3 right parenthesis minus 4 over denominator square root of 6 squared plus left parenthesis negative 3 right parenthesis squared plus 2 squared end root end fraction close vertical bar space space space rightwards double arrow straight d space equals space open vertical bar fraction numerator 12 minus 15 minus 6 minus 4 over denominator square root of 36 plus 9 plus 4 end root end fraction close vertical bar space space space space rightwards double arrow straight d space equals space open vertical bar fraction numerator negative 13 over denominator square root of 49 end fraction close vertical bar space space space space rightwards double arrow straight d space equals space 13 over 7 units$

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Integrals

Find the distance of a point (2, 5, −3) from the plane $straight r with rightwards arrow on top. open parentheses 6 straight i with hat on top minus 3 straight j with hat on top plus 2 straight k with hat on top close parentheses space equals 4.$

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