1. Find the line through the point [ 1 ,  3 ,  3 ] and parallel to the vector [ 1 ,  4 ,  2 ].

    [ x ,  y ,  z ]  = [ 1 + t ,  3 + 5t ,  3 + 3t ] [ x ,  y ,  z ]  = [ 1 + t ,  3 + 4t ,  3 + 2t ] [ x ,  y ,  z ]  = [ 2 + t ,  3 + 4t ,  4 + 2t ] [ x ,  y ,  z ]  = [ 2 + t ,  3 + 5t ,  4 + 3t ]

  2. Find the line through the points [  − 1 ,  0 ,  0 ] and [ 2 ,  1 ,   − 1 ] .

    [ x ,  y ,  z ]  = [  − 1 + 3t ,  2t ,   − t ] [ x ,  y ,  z ]  = [ 3t ,  t ,  1 − t ] [ x ,  y ,  z ]  = [ 3t ,  2t ,  1 − t ] [ x ,  y ,  z ]  = [  − 1 + 3t ,  t ,   − t ]

  3. Find the line through the point [  − 3 ,   − 2 ,   − 3 ] and perpendicular to the plane 1 + x + 2z = 0.

    [ x ,  y ,  z ]  = [  − 3 + t ,   − 2 ,   − 3 + 2t ] [ x ,  y ,  z ]  = [  − 3 + t ,   − 2 + t ,   − 3 + 3t ] [ x ,  y ,  z ]  = [  − 2 + t ,   − 2 + t ,   − 3 + 3t ] [ x ,  y ,  z ]  = [  − 2 + t ,   − 2 ,   − 3 + 2t ]

  4. Find the line of intersection of the planes −x + 3y = 0 and −16 − 3x − y + 2z = 0 using the point [  − 3 ,   − 1 ,  3 ] of intersection.

    [ x ,  y ,  z ]  = [  − 3 − t ,   − 1 + 3t ,  3 ] [ x ,  y ,  z ]  = [  − 3 − 3t ,   − 1 − t ,  3 + 2t ] $\displaystyle \left[ x , y , z \right] =\left[ -3+6\,t , -1+2\,t , 3+10\,t \right] $ $\displaystyle \left[ x , y , z \right] =\left[ -3+6\,t , -1+3\,t , 3+11\,t \right] $

  5. Find the plane through the point [ 3 ,  2 ,   − 4 ] and perpendicular to the vector [ 4 ,   − 1 ,   − 1 ].

    −18 + 4x − y − z = 0 −20 + 4x − z = 0 −16 + 4x − z = 0 −14 + 4x − y − z = 0

  6. Find the plane through the points [  − 2 ,  0 ,   − 3 ], [  − 2 ,  1 ,   − 4 ], and [  − 4 ,   − 1 ,   − 2 ].

    4 + 3y + 2z = 0 6 + 3y + 2z = 0 4 + 2y + 2z = 0 6 + 2y + 2z = 0

  7. Find the plane through the point [  − 2 ,  1 ,   − 3 ] and containing the line [ x ,  y ,  z ]  = [ 1 + t ,  1 − 4t ,   − 2 + 2t ].

    23 − 4x + 5y + 12z = 0 27 − 4x + 5y + 12z = 0 26 − 4x + 6y + 12z = 0 22 − 4x + 6y + 12z = 0

  8. Find the distance between the point [ 1 ,  3 ,   − 1 ] and the planes −3 − 3x + 2y + 4z = 0.

    0 $\displaystyle \frac{3}{\sqrt{29}}$ $\displaystyle \frac{1}{\sqrt{43}}$ $\displaystyle \frac{4}{\sqrt{29}}$

  9. Find the distance between the planes 16 − 3x − 2y − 2z = 0 and −13 − 3x − 2y − 2z = 0.

    $\displaystyle \frac{25}{\sqrt{14}}$ $\displaystyle \frac{29}{\sqrt{17}}$ $\displaystyle \frac{24}{\sqrt{14}}$ $\displaystyle \frac{30}{\sqrt{17}}$

  10. Find the tangent line to $\mathbf{r}(s) = \left[ s^2 , s , s^3 \right] $ at $s = 1$.

    [ x ,  y ,  z ]  = [ 1 + 2t ,  1 + t ,  1 + 3t ] [ x ,  y ,  z ]  = [ 1 + 2t ,  1 + 2t ,  1 + 3t ] [ x ,  y ,  z ]  = [ 2 + 2t ,  1 + 2t ,  1 + 3t ] [ x ,  y ,  z ]  = [ 2 + 2t ,  1 + t ,  1 + 3t ]

  11. Let $\mathbf{r}(t) = \mathbf{u}(t)\times\mathbf{v}(t)$. Suppose that $\mathbf{u}( -1) = \left[ -2 , -3 , 0 \right] $, $\mathbf{v}( -1) = \left[ 2 , -2 , -3 \right] $, $\mathbf{u}'( -1) = \left[ -1 , 1 , 1 \right] $, and $\mathbf{v}'( -1) = \left[ 2 , 1 , 0 \right] $. Then find $\mathbf{r}'( -1)$.

    [  − 1 ,   − 1 ,  4 ] [  − 2 ,   − 1 ,  3 ] [  − 2 ,  1 ,  5 ] [  − 3 ,  1 ,  4 ]

  12. Let $\mathbf{r}(t) = \mathbf{u}(t)\cdot\mathbf{v}(t)$. Suppose that $\mathbf{u}( 1) = \left[ -3 , 2 , 1 \right] $, $\mathbf{v}( 1) = \left[ -1 , 2 , -2 \right] $, $\mathbf{u}'( 1) = \left[ -2 , -2 , 1 \right] $, and $\mathbf{v}'( 1) = \left[ 1 , 0 , 1 \right] $. Then find $\mathbf{r}'( 1)$.

    −6 −5 −5 −8


Department of Mathematics
Last modified: 2026-03-17