Find the intersection of the line and plane: x + 2 y + 2 z = 3 r ( t ) =&lt; 2 , − 3 , 3 &gt; + t &lt; 1 , 3 , − 3 &gt;

Question

Find the intersection of the line and plane:  x  +  2  y  +  2  z  =  3    r  (  t  )  =&amp;lt;  2  ,  −  3  ,  3  &amp;gt;  +  t  &amp;lt;  1  ,  3  ,  −  3  &amp;gt;

Raina Russo · Accepted Answer

Equation of plane is:  x  +  2  y  +  2  z  =  3and Equation of Straight line is:  r  (  t  )  =&amp;lt;  2  ,  −  3  ,  3  &amp;gt;  +  t  &amp;lt;  1  ,  3  ,  −  3  &amp;gt;The cantesian equation of straight line            x      −      2        1    =            y      +      3        3    =            7      −      3              −      3        =  t    x  =  2  +  t     y  =  3  t  −  3     z  =  3  −  3  tLet the point of intersection be  P  (  2  +  t  ,  3  t  −  3  ,  3  −  3  t  )So:  P  (  2  +  t  ,  3  t  −  3  ,  3  −  3  t  )  Since the point P lies on the planeFrom   x  +  2  y  +  27  =  3    ⇒  (  2  +  t  )  +  2  (  3  t  −  3  )  +  2  (  3  −  3  t  )  =  3    ⇒  2  +  t  +  6  t  −  6  +  6  −  6  t  =  3    t  =  3  −  2    t  =  1put   t  =  1    x  =  2  +  1    =  3The point of intersection  P  =  (  3  ,  0  ,  0  )

Find the intersection of the line and plane: x+2y+2z=3, r(t)=<2,-3,3>+t<1,3,-3>

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