right triangle ABC w/ AB = 4x, BC = 12, AD = (2/3)AC. Find...

[kuben]

Need help with this question on vectors:

\(\displaystyle \mbox{In a right-angled tria}\)\(\displaystyle \mbox{ngle }\, ABC\, \mbox{ with the right an}\)\(\displaystyle \mbox{gle at }\, A,\)

\(\displaystyle \overrightarrow{AB}\, =\, 4x,\, \overrightarrow{BC}\, =\, 12y,\, \mbox{ and }\, AD\, =\, \dfrac{2}{3}\, AC.\)

\(\displaystyle \mbox{Point }\, D\, \mbox{ is on the vertical from }\, A\, \mbox{ to }\, C.\)

\(\displaystyle \mbox{(a) Find }\, \overrightarrow{BD}\, \mbox{ in terms of }\, x\, \mbox{ and }\, y.\)

\(\displaystyle \mbox{(b) Find }\, \lvert \overrightarrow{BD} \rvert\, \mbox{ if }\, \lvert x \rvert \, =\, 3\, \mbox{ and }\, \lvert y \rvert = 2.\)

 

[stapel]

Need help with this question on vectors:

\(\displaystyle \mbox{In a right-angled tria}\)\(\displaystyle \mbox{ngle }\, ABC\, \mbox{ with the right an}\)\(\displaystyle \mbox{gle at }\, A,\)

\(\displaystyle \overrightarrow{AB}\, =\, 4x,\, \overrightarrow{BC}\, =\, 12y,\, \mbox{ and }\, AD\, =\, \dfrac{2}{3}\, AC.\)

\(\displaystyle \mbox{Point }\, D\, \mbox{ is on the vertical from }\, A\, \mbox{ to }\, C.\)

\(\displaystyle \mbox{(a) Find }\, \overrightarrow{BD}\, \mbox{ in terms of }\, x\, \mbox{ and }\, y.\)

\(\displaystyle \mbox{(b) Find }\, \lvert \overrightarrow{BD} \rvert\, \mbox{ if }\, \lvert x \rvert \, =\, 3\, \mbox{ and }\, \lvert y \rvert = 2.\)

Where are you stuck? They gave you a right-angled triangle, so you applied the Pythagorean Theorem to get an expression for the length of AC in terms of x and y. You applied the relation they gave you to obtain an expression for the length of AD in terms of x and y. You applied the Pythagorean Theorem again to find the length of BD. You plugged the given values into your expression for the length. And... then what?

Please be complete. Thank you!