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Vector Algebra

Question
CBSEENMA12035696
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If the cartesian equations of a line are fraction numerator 3 minus straight x over denominator 5 end fraction equals fraction numerator straight y plus 4 over denominator 7 end fraction equals fraction numerator 2 straight z minus 6 over denominator 4 end fraction comma write the vector equation for the line. 

Solution

Given that the cartesian equation of the line as 
fraction numerator 3 minus straight x over denominator 5 end fraction equals fraction numerator straight y plus 4 over denominator 7 end fraction equals fraction numerator 2 straight z minus 6 over denominator 4 end fraction That space is comma fraction numerator negative left parenthesis straight x minus 3 right parenthesis over denominator 5 end fraction equals fraction numerator straight y minus left parenthesis negative 4 right parenthesis over denominator 7 end fraction equals fraction numerator 2 left parenthesis straight z minus 3 right parenthesis over denominator 4 end fraction rightwards double arrow fraction numerator straight x minus 3 over denominator negative 5 end fraction equals fraction numerator straight y minus left parenthesis negative 4 right parenthesis over denominator 7 end fraction equals fraction numerator straight z minus 3 over denominator 2 end fraction equals straight lambda
Any point on the line is of the form:
negative 5 straight lambda plus 3 comma space space 7 straight lambda minus 4 comma space space 2 straight lambda plus 3
Thus, the vector equation is of the form:
straight r with rightwards arrow on top space equals straight a with rightwards arrow on top plus straight lambda straight b with rightwards arrow on top comma space where space straight a with rightwards arrow on top space is space the space position space vector space of space any space point on space the space line space and space straight b with rightwards arrow on top space is space the space vector space parallel space to space the space line. space Therefore comma space the space vector space equation space is straight r with rightwards arrow on top equals space left parenthesis negative 5 straight lambda plus 3 right parenthesis straight i with hat on top plus open parentheses 7 straight lambda minus 4 close parentheses straight j with hat on top space plus space left parenthesis 2 straight lambda plus 3 right parenthesis straight k with hat on top space space rightwards double arrow space straight r with rightwards arrow on top equals negative 5 straight lambda straight i with hat on top plus 7 straight lambda straight j with hat on top plus 2 straight lambda straight k with hat on top plus 3 straight i with hat on top minus 4 straight j with hat on top plus 3 straight k with hat on top space space space space rightwards double arrow straight r with rightwards arrow on top equals 3 straight i with hat on top minus 4 straight j with hat on top plus 3 straight k with hat on top plus straight lambda open parentheses negative 5 straight i with hat on top plus 7 straight j with hat on top plus 2 straight k with hat on top close parentheses

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