Porisma

Worked solution

Find a normal unit vector and direction cosines

AA, BB and CC are points (0,1,2)(0,1,2), (3,2,1)(3,2,1) and (1,−1,0)(1,-1,0) respectively.

(i) Find the unit vector perpendicular to both AB→\overrightarrow{AB} and AC→\overrightarrow{AC}.

(ii) Find the cosine of the angles that OB→\overrightarrow{OB} makes with the x-, y- and z-axis.

CJC · Discussion · Question 7

Answer

(i) ±1310(−4,5,−7)\pm\dfrac 1{3\sqrt{10}}(-4,5,-7); (ii) 3/143/\sqrt{14}, 2/142/\sqrt{14}, 1/141/\sqrt{14}.

(i) Form the two displacement vectors

AB→=(31−1),AC→=(1−2−2).\overrightarrow{AB}=\begin{pmatrix}3\\1\\-1\end{pmatrix},\qquad\overrightarrow{AC}=\begin{pmatrix}1\\-2\\-2\end{pmatrix}.

Their cross product is AB→×AC→=(−4,5,−7)\overrightarrow{AB}\times\overrightarrow{AC}=(-4,5,-7). Its squared norm is 16+25+49=9016+25+49=90.

How did this part go?

(i) Normalise the perpendicular vector

One unit vector is

1310(−45−7).\frac 1{3\sqrt{10}}\begin{pmatrix}-4\\5\\-7\end{pmatrix}.

Its dot products with AB and AC are both zero. Its negative is also a perpendicular unit vector; the source prints this one orientation.

How did this part go?

(ii) Compute the three direction cosines

OB→=(3,2,1)\overrightarrow{OB}=(3,2,1) has magnitude 14\sqrt{14}. Dotting its unit vector with the three coordinate unit vectors gives

cos⁡θx=314,cos⁡θy=214,cos⁡θz=114.\cos\theta_x=\frac 3{\sqrt{14}},\quad\cos\theta_y=\frac 2{\sqrt{14}},\quad\cos\theta_z=\frac 1{\sqrt{14}}.

Their squared values sum to 1, a useful check.

How did this part go?

Common mistakes

The phrase “the unit vector” does not select an orientation. Reversing the cross-product order gives the other valid sign.

Solution notes

Both source asks, cross-product components, normalization and all three axis dot products retained. No source diagram printed for this question.

Read the hint