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 A  symmetrical  form  of  the  line  of  intersection of the planes   x=ay+b \: and \: z=cy+d  is

  • Option 1)

    \frac{x-b}{a}=\frac{y-1}{1}=\frac{z-d}{c}

  • Option 2)

    \frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}

  • Option 3)

    \frac{x-a}{b}=\frac{y-0}{1}=\frac{z-c}{d}

  • Option 4)

    \frac{x-b-a}{b}=\frac{y-1}{0}=\frac{z-d-c}{d}

 

Answers (1)

best_answer

As we learnt in

Cartesian eqution of a line -

The equation of a line passing through two points A\left ( x_{0},y_{0},z_{0} \right )and parallel to vector having direction ratios as \left ( a,b,c \right )is given by

\frac{x-x_{0}}{a}= \frac{y-y_{0}}{b}= \frac{z-z_{0}}{c}

The equation of a line passing through two points A\left ( x_{1},y_{1},z_{1} \right )\, and \, B\left ( x_{2},y_{2},z_{2} \right ) is given by

\frac{x-x_{1}}{x_{2}-x_{1}}= \frac{y-y}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}

- wherein

 

 x-ay-b=0 and cy-z+d=0

Normal vector will be 

\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1& -a& 0\\ 0&c &-1 \end{vmatrix}

Vector = a\hat{i}+\hat{j}+c\hat{k}

So DRs are (a, 1, c)

Put y=1

x=a+b and z=c+d

So live becomes 

\frac{x-\left ( a+b \right )}{a}= \frac{y-1}{1} =\frac{z-\left ( c+d \right )}{c}


Option 1)

\frac{x-b}{a}=\frac{y-1}{1}=\frac{z-d}{c}

This option is incorrect

Option 2)

\frac{x-b-a}{a}=\frac{y-1}{1}=\frac{z-d-c}{c}

This option is correct

Option 3)

\frac{x-a}{b}=\frac{y-0}{1}=\frac{z-c}{d}

This option is incorrect

Option 4)

\frac{x-b-a}{b}=\frac{y-1}{0}=\frac{z-d-c}{d}

This option is incorrect

Posted by

divya.saini

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