N is the midpoint of AD, and the segment BN…

N is the midpoint of AD, and the segment BN meets the diagonal AC at G.

Question 55

HardOriginal question

Three vertices of a parallelogram $ABCD$ , taken in order, are $A(1, 2, -1)$ , $B(2, -1, 0)$ and $C(5, 4, 3)$ .

$N$ is the midpoint of $AD$ , and the segment $BN$ meets the diagonal $AC$ at $G$ . What are the coordinates of $G$ ?

(3, 3, 1)

Correct answerYour answer

(\frac{11}{3}, \frac{10}{3}, \frac{5}{3})

Correct answerYour answer

(\frac{5}{2}, \frac{9}{2}, \frac{1}{2})

Correct answerYour answer

(\frac{7}{3}, \frac{8}{3}, \frac{1}{3})

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

The diagonals meet at the midpoint $M$ of $BD$ , so $AM$ and $BN$ are both medians of triangle $ABD$ ; they meet at its centroid.
$G = \frac{\vec{a} + \vec{b} + \vec{d}}{3} = \frac{(7, 8, 1)}{3} = (\frac{7}{3}, \frac{8}{3}, \frac{1}{3})$ , a third of the way from $A$ to $C$ .

Worth noting

$(3, 3, 1)$ is where the diagonals meet; $BN$ crosses $AC$ before that, at the centroid of triangle $ABD$ .

Ten questions from Vectors, at your own pace.

Ten questions from Vectors, at your own pace.