Exam-Style Problem

9231 P11 - Jun 2017 - Q12O - 14 marks

6242

The position vectors of the points \(A\), \(B\), \(C\), \(D\) are \(\mathbf{i}+\mathbf{j}+3\mathbf{k}\), \(3\mathbf{i}-\mathbf{j}+5\mathbf{k}\), \(3\mathbf{i}-\mathbf{j}+\mathbf{k}\), and \(5\mathbf{i}-5\mathbf{j}+\alpha\mathbf{k}\), respectively, where \(\alpha\) is a positive integer.

It is given that the shortest distance between the line \(AB\) and the line \(CD\) is equal to \(2\sqrt2\).

(i) Show that the possible values of \(\alpha\) are \(3\) and \(5\).

(ii) Using \(\alpha=3\), find the shortest distance of the point \(D\) from the line \(AC\), giving your answer correct to 3 significant figures.

(iii) Using \(\alpha=3\), find the acute angle between the planes \(ABC\) and \(ABD\), giving your answer correct to 3 significant figures.

Solution Checked by expert

Answer: (i) \(\alpha=3\) or \(\alpha=5\).

(ii) The shortest distance is \(\sqrt{\frac{56}{3}}=4.32\) to 3 significant figures.

(iii) The acute angle is \(19.1^\circ\), or \(0.333\) radians.

Use coordinates \(A=(1,1,3)\), \(B=(3,-1,5)\), \(C=(3,-1,1)\), and \(D=(5,-5,\alpha)\).

Then

\(\overrightarrow{AB}=(2,-2,2)\), \(\overrightarrow{AC}=(2,-2,-2)\), and \(\overrightarrow{CD}=(2,-4,\alpha-1)\).

For the distance between the skew lines \(AB\) and \(CD\), use

\(d=\frac{|\overrightarrow{AC}\cdot(\overrightarrow{AB}\times\overrightarrow{CD})|}{|\overrightarrow{AB}\times\overrightarrow{CD}|}\).

It is convenient to use \((1,-1,1)\) instead of \(\overrightarrow{AB}\), since it is a scalar multiple. Then

\((1,-1,1)\times(2,-4,\alpha-1)=(5-\alpha,3-\alpha,-2)\).

Equivalently, take \((\alpha-5,\alpha-3,2)\) as the normal direction.

Now

\(\overrightarrow{AC}\cdot(\alpha-5,\alpha-3,2)=2(\alpha-5)-2(\alpha-3)-4=-8\).

Also

\(|(\alpha-5,\alpha-3,2)|=\sqrt{(\alpha-5)^2+(\alpha-3)^2+4}=\sqrt{2\alpha^2-16\alpha+38}\).

Since the distance is \(2\sqrt2\),

\(\frac{8}{\sqrt{2\alpha^2-16\alpha+38}}=2\sqrt2\).

Squaring gives

\(64=8(2\alpha^2-16\alpha+38)\), so

\(2\alpha^2-16\alpha+38=8\).

Thus

\(\alpha^2-8\alpha+15=0\), giving

\((\alpha-3)(\alpha-5)=0\).

Hence \(\alpha=3\) or \(\alpha=5\).

Now take \(\alpha=3\), so \(D=(5,-5,3)\). Then

\(\overrightarrow{AD}=(4,-6,0)\) and \(\overrightarrow{AC}=(2,-2,-2)\).

The shortest distance from \(D\) to the line \(AC\) is

\(\frac{|\overrightarrow{AD}\times\overrightarrow{AC}|}{|\overrightarrow{AC}|}\).

Using \((1,-1,-1)\) in place of \(\overrightarrow{AC}\),

\(\overrightarrow{AD}\times(1,-1,-1)=(6,4,2)\), so its magnitude is \(\sqrt{36+16+4}=\sqrt{56}\).

Since \(|(1,-1,-1)|=\sqrt3\), the distance is

\(\sqrt{\frac{56}{3}}=4.32\) to 3 significant figures.

For the angle between the planes, find normal vectors.

For plane \(ABC\), a normal vector is

\(\mathbf{n}_1=(1,-1,-1)\times(1,-1,1)=(-1,-1,0)\).

For plane \(ABD\), using \((1,-1,1)\) and \((2,-3,0)\), a normal vector is

\(\mathbf{n}_2=(1,-1,1)\times(2,-3,0)=(3,2,-1)\).

The acute angle between the planes is the acute angle between their normals, so

\(\cos\theta=\left|\frac{\mathbf{n}_1\cdot\mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|}\right|=\left|\frac{-3-2+0}{\sqrt2\sqrt{14}}\right|=\frac{5}{\sqrt{28}}\).

Therefore

\(\theta=19.1^\circ\), or \(0.333\) radians.