Image of the point P with position vector

\vec{7i}-\vec{j}+\vec{2k}\: in\: the \:line\: whose\:vector\:equation\: is,

\vec{r}=\vec{9i}+\vec{5j}+\vec{5k} +5\vec{K}+\lambda (\vec{i}+\vec{3j}+\vec{5k}) has the position vector:

  • Option 1)


  • Option 2)


  • Option 3)


  • Option 4)



Answers (1)

As learnt in

Perpendicular distance of a point from a line, foot of perpendicular and image (Cartesian form ) - - wherein

L be the foot of perpendicular drawn from P(\alpha ,\beta ,\gamma ) on the line \frac{x-x_{1}}{a}= \frac{y-y_{1}}{b},\frac{z-z_{1}}{c}

Co-ordinates of L will be \left ( x_{1} +\lambda a,y_{1}+\lambda b,z_{1}+\lambda c\right )

DR's of PL will be \left ( x_{1} +\lambda a-\alpha ,y_{1}+\lambda b-\beta ,z_{1}+\lambda c-\gamma \right )

PL is perpendicular to line,\left ( x_{1} +\lambda a-\alpha \right )a+\left ( y_{1}+\lambda b-\beta \right )b+\left ( z_{1}+\lambda c-\gamma \right )c= 0

Find \gamma and put in coordinates of L Distance between P and L is perpendicular distance Image Q can be obtained by applying mid point formula in P and Q and equating it to L obtained.



 ((7-\alpha )\hat{i}-(1+\beta )\hat{j}+(2-\gamma )\hat{k})\cdot (\hat{i}+3\hat{j}+5\hat{k}) =0  

\Rightarrow (7-\alpha ) - 3(1+\beta )+5(2-\gamma )=0

\Rightarrow 14-\alpha -3\beta -5\gamma =0 - (i)

We have equation of the line in cartesian form:

\frac{x-9}{1}=\frac{y-5}{3} = \frac{z-5}{5}

x=\frac{\alpha +7}{2}; y= \frac{\beta -1}{2}; z=\frac{y+z}{2} 

lie on the given line. 

So, \frac{\alpha +7}{2}-9 = \frac{\frac{\beta -1}{2}-5}{3} = \frac{\frac{\gamma +2}{2}-5}{5}=k

\Rightarrow \alpha +7 = 2 (9+k);

\beta -1 = 2(3k+5);

\gamma +2= 2 (5k+5)

So, \alpha =2k+11; \beta =6k+11; \gamma =10k+8

Putting iv (i),


                               = 0

\Rightarrow 14-70k-84=0

\Rightarrow 70k=-70\Rightarrow k=-1

So, \alpha =9; \beta =5; \gamma =-2

Option 1)


This solution is incorrect

Option 2)


This solution is correct

Option 3)


This solution is incorrect

Option 4)


This solution is incorrect

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