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  • Please,please help me - Three Dimensional Geometry - JEE Main

If the angle between the lines,  x/2=y/2= z/1

and \frac{5-x}{-2}= \frac{7y-14}{p}= \frac{z-3}{4}

is \cos ^{\left (-1 \right )}\frac{2}{3}

then p is equal to :

  • Option 1)

    7/2

  • Option 2)

    2/7

  • Option 3)

    -7/4

  • Option 4)

    -4/7

 

Answers (1)

best_answer

As we have learned

Angle between two lines (cartesian form ) -

Let the two lines be \frac{x-x_{1}}{a_{1}}= \frac{y-y_{1}}{b_{1}}= \frac{z-z_{1}}{c_{1}} and \frac{x-x_{2}}{a_{2}}= \frac{y-y_{2}}{b_{2}}= \frac{z-z_{2}}{c_{2}}

The angle between them is given by

\cos \Theta = \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a{_{1}}^{2}+b{_{1}}^{2}+c{_{1}}^{2}}\sqrt{a{_{2}}^{2}+b{_{2}}^{2}+c{_{2}}^{2}}}

-

 

 \frac{x}{2}=\frac{y}{2}=\frac{z}{2}

\frac{x-5}{2}=\frac{y-2}{p/7}=\frac{z-13}{4}

\cos \Theta = \frac{a_{1a_{2}+b_{1}b_{2}+c_{1}c_{2}}}{\sqrt{(a_{1}^{2}+b_{1}^{2}+c_{1}^{2})\sqrt{(a_{2}^{2}+b_{2}^{2}+c_{2}^{2})}}}

2/3=\frac{4+2p/7+4}{3\times \sqrt(20+\frac{p^{2}}{49})}

\sqrt{{20}+\frac{p^{2}}{49}}= 4+ p/7

p= 7/2

 

 

 

 


Option 1)

7/2

This is correct 

Option 2)

2/7

This is incorrect 

Option 3)

-7/4

This is incorrect 

Option 4)

-4/7

This is incorrect 

Posted by

Himanshu

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