A line with equation r = a + λ d → meets plane π with equation r . n ^ = k at point P. Point Q lies in π and is the foot of the perpendicular from A to π. Find the direction vector of line PQ.By solving ( a + λ d → ) . n ^ = k, I was able to find the position vector of P. Then by finding the intersection of line AQ and plane I was able to find the position vector of Q and hence the direction vector PQ.However, the answer can be found simply by finding ( n ^ × d → ) × n ^ where × is cross-product. I don't understand why.Here's what I know : The cross-product of 2 vectors gives a 3rd vector perpendicular to the 2 vectors. Line PQ lies on plane so direction vector PQ ⊥ n ^ . Also, AQ is parallel to n ^ .The first part w = ( n ^ × d → ) gives a vector perpendicular to line and parallel to plane. Won't w × n give a vector perpendicular to the plane again? I can't understand the geometric interpretation of ( n ^ × d → ) × n ^ .

Question

A line with equation   r  =  a  +  λ            d      →        meets plane π with equation   r  .            n      ^        =  k at point P. Point Q lies in π and is the foot of the perpendicular from A to π. Find the direction vector of line PQ.By solving   (  a  +  λ            d      →        )  .            n      ^        =  k, I was able to find the position vector of P. Then by finding the intersection of line AQ and plane I was able to find the position vector of Q and hence the direction vector PQ.However, the answer can be found simply by finding   (            n      ^        ×            d      →        )  ×            n      ^       where × is cross-product. I don&#039;t understand why.Here&#039;s what I know : The cross-product of 2 vectors gives a 3rd vector perpendicular to the 2 vectors. Line PQ lies on plane so direction vector PQ   ⊥            n      ^      . Also, AQ is parallel to             n      ^      .The first part   w  =  (            n      ^        ×            d      →        ) gives a vector perpendicular to line and parallel to plane. Won&#039;t   w  ×  n give a vector perpendicular to the plane again? I can&#039;t understand the geometric interpretation of   (            n      ^        ×            d      →        )  ×            n      ^      .

Alaina Mcintosh · Accepted Answer

Imagine this line   r  =  a  +  λ            d      →        is intersecting plane   π  :            r      →        .            n      ^        =  kThe cross product of             n      ^        ×            d      ^        =                    u        1            ^       this                     u        1            ^       will be the unit vector normal to the plane of line containing line   r  =  a  +  λ            d      →        and plane   π  :            r      →        .            n      ^        =  kNow, we take the cross product of                     u        1            ^       and             n      ^      :                     u        2            ^        =                    u        1            ^        ×            n      ^       will be the required unit vector.Note: Here the                     u        2            ^       depends on the unit vector             d      ^       I mean                     u        ^              2    =                    u        1            ^        ×            n      ^       or                     u        2            ^        =            n      ^        ×                    u        1            ^      Find the direction vector of line PQ.Here, I used unit vectors only as you were interested in the direction of                     P        Q            →      ;

Makayla Eaton · Accepted Answer

The segment PQ lies in a plane that is spanned by the perpendicular to the plane n and the direction vector d, so the normal to this plane is  n × d . But PQ also lies in the plane whose normal is n, hence the direction vector of PQ must be along the vector  ( n × d ) × n &quot;

A line with equation r=a+lambda vec(d) meets plane pi with equation r. hat(n)=k at point P. Point Q lies in π and is the foot of the perpendicular from A to pi. Find the direction vector of line PQ.

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