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Integrals

Question
CBSEENMA12035653
Wired Faculty App

Find the equation of the plane which contains the line of intersection of the planes
straight r with rightwards arrow on top. left parenthesis straight i with hat on top minus 2 straight j with hat on top plus 3 straight k with hat on top right parenthesis minus 4 space equals space 0 and straight r with rightwards arrow on top. left parenthesis negative 2 straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis plus 5 space space equals space 0

and whose intercept on the x-axis is equal to that of on y-axis.

Solution
stack straight r. with rightwards arrow on top left parenthesis straight i with hat on top minus 2 straight j with hat on top plus 3 straight k with hat on top right parenthesis minus 4 space equals 0 straight r with rightwards arrow on top. left parenthesis negative 2 straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis plus 5 space equals 0 straight r with rightwards arrow on top space left parenthesis straight i with hat on top minus 2 straight j with hat on top plus 3 straight k with hat on top right parenthesis plus straight lambda open curly brackets straight r with rightwards arrow on top. left parenthesis negative 2 straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis close curly brackets minus 4 plus 5 straight lambda space equals 0 rightwards double arrow space straight r with rightwards arrow on top left square bracket left parenthesis 1 minus 2 straight lambda right parenthesis straight i with hat on top plus left parenthesis negative 2 plus straight lambda right parenthesis straight j with hat on top space plus left parenthesis 3 plus straight lambda right parenthesis straight k with hat on top right square bracket minus 4 plus 5 straight lambda equals 0 left parenthesis 1 minus 2 straight lambda right parenthesis space straight x space plus left parenthesis negative 2 plus straight lambda right parenthesis straight y plus left parenthesis 3 plus straight lambda right parenthesis straight z space equals negative 5 straight lambda plus 4 rightwards double arrow space fraction numerator straight x over denominator begin display style fraction numerator negative 5 straight lambda plus 4 over denominator 1 minus 2 straight lambda end fraction end style end fraction plus fraction numerator straight y over denominator begin display style fraction numerator negative 5 straight lambda plus 4 over denominator negative 2 plus straight lambda end fraction end style end fraction plus fraction numerator straight z over denominator begin display style fraction numerator negative 5 straight lambda plus 4 over denominator 3 plus straight lambda end fraction end style end fraction equals 1 therefore space fraction numerator negative 5 straight lambda plus 4 over denominator 1 minus 2 straight lambda end fraction space equals fraction numerator negative 5 straight lambda plus 4 over denominator negative 2 plus straight lambda end fraction rightwards double arrow space 1 minus 2 straight lambda space equals negative 2 plus straight lambda rightwards double arrow space minus 3 straight lambda space equals space minus 3 rightwards double arrow space straight lambda space equals 1 therefore space equation space of space the space required space plane minus straight x minus straight y plus 4 straight z equals 1 straight x plus straight y minus 4 straight z minus 1 space equals space 0 vector space equation space of space the space required space plane equals straight r with rightwards arrow on top. left parenthesis straight i with hat on top plus straight j with hat on top minus 4 straight k with hat on top right parenthesis minus 1 space equals 0

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