# A triangle has sides formed by vectors PA=(2,7,3) and PB=(6,2,2). The point P=(1,3,2). Find the coordinates of vertices.

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You need to find the coordinate of vertices A and B.

You know that the vector `barPA = (x_P - x_A)bari + (y_P - y_A)barj +(z_P - z_A)bark`

The problem provides the information that `barPA = 2bari + 7barj + 3bark` Equating the coefficients of `bari,barj,bark` yields:

`x_P - x_A = 2`

`y_P - y_A = 7`

`z_P - z_A = 3`

You need to substitute 1 for `x_P` , 3 for `y_P` , 2 for `z_P` in equations above such that:

`1- x_A = 2 =gt x_A = -1`

`3 - y_A = 7 =gt y_A = -4`

`2 - z_A = 3 =gt z_A = -1`

Thus, evaluating the coordinates of A vertex yields (-1,-4,-1).

You know that the vector `barPA = (x_P - x_B)bari + (y_P - y_B)barj +(z_P - z_B)bark`

The problem provides the information that `barPB = 6bari + 2barj + 2bark` Equating the coefficients of `bari` ,`bar j` ,`bark` yields:

`x_P - x_B = 6`

`y_P - y_B = 2`

`z_P - z_B = 2`

You need to substitute 1 for `x_P` , 3 for `y_P` , 2 for `z_P` in equations above such that:

`1- x_B = 6 =gt x_B = -5`

`3- y_B = 2 =gt y_B = 1`

`2- z_B = 2 =gt z_B = 0`

Thus, evaluating the coordinates of B vertex yields (-5,1,0).

**Hence, evaluating the vertices of triangle yields (-1,-4,-1) coordinates for A and (-5,1,0) coordinates for B.**