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The planes $\pi_1$ and $\pi_2$ have vector equations: $$\pi_1: r=\lambda_1(i+j-k)+\mu_1(2i-j+k)$$ $$\pi_2: r=\lambda_2(i+2j+k)+\mu_2(3i+j-k)$$ $i.$ The line $l$ passes through the point with position vector $4i+5j+6k$ and is parallel to both $\pi_1$ and $\pi_2$. Find a vector equation for $l$.

This is what I know: $$l \parallel \pi_1, \pi_2 \implies l \perp n_1, n_2 \implies d = n_1\times n_2;\quad l:r=(4i+5j+6k)+\lambda d$$ However, that method involves 3 cross-products, which according to the examination report was an 'expeditious' solution which was also prone to 'lack of accuracy'. Rightfully so, I also often make small errors with sign changes leading me to the wrong answer, so if there is a more efficient way of working I'd like to know what that is.

Q1 How can I find the equation for line $l$ without converting the plane equations to cartesian form?

$ii.$ Find also the shortest distance between $l$ and the line of intersection of $\pi_1$ and $\pi_2$.

Three methods are described to solve the second part. However, I didn't understand this one:

The determination of the shortest distance was perceived by most to be the shortest distance of the given point $4i+5j+6k$ to the line of intersection of the planes. Thus they continued by evaluating the vector product $(4i+5j+6k)\times (3i+j-k)$ to obtain $-11i+22j-11k$. The required minimum distance is then given immediately by $p = \frac{|-11i+22j-11k|}{\sqrt{11}} = \sqrt{66}$

Q2 Why does the cross-product of the given point and the direction vector of the line of intersection of the planes get to the correct answer?

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1 Answer 1

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1) One method you can use to find line $L_1$ is to make equations in $x, y, z$ for each plane and solve them, instead of finding the cross product: $$x=\lambda_1+2\mu_1, y=\lambda_1-\mu_1, z=-\lambda_1+\mu_1$$ $$x=\lambda_2+3\mu_2, y=\lambda_2+\mu_2, z=\lambda_2-\mu_2$$

What we'll do is find the line of intersection; this line is clearly parallel to both planes so we can use its direction for the line through $(4,5,6)$ that we want.

Since the line of intersection has to lie on both planes, it will satisfy both sets of equations. So equating them,

$$\lambda_1+2\mu_1 = \lambda_2+3\mu_2$$ $$\lambda_1-\mu_1 = 2\lambda_2+\mu_2$$ $$-\lambda_1+\mu_1 = \lambda_2-\mu_2$$

Adding the last two equations, we get, $$\lambda_2 = 0$$

So, $$x=3\mu_2, y=\mu_2, z=-\mu_2$$

This is a parametrisation of the solution which is the line. So obviously $L_1$ must be: $$L_1: (4,5,6) + \mu_2(3,1,-1)$$

2) Notice that $(0,0,0)$ lies on the line of intersection $L_2$. So the difference vector $v$ between the point $(4,5,6)$ on the first line and $(0,0,0)$ on the second is $v=(4,5,6)$. The distance is really the length of the projection of $v$ on the perpendicular joining the two lines. This is $|v|\sin \theta$ where $\theta$ is the angle between line and plane.

Expressing $|v|\sin\theta$ as $$|v|\sin\theta=\frac{|v||w|\sin\theta}{|w|} = \frac{|v\times w|}{|w|}$$ (where $w$ is the direction vector of $L_2$) gives you the answer.

Here taking the cross product of the position vector of the given point with the direction vector of $L_2$ depended on $(0,0,0)$ lying on $L_2$; otherwise you would need to take the difference vector of the given point and any point on $L_2$.

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I came across another similar question which taught me the method for part i. Almost the same as yours with slight difference in technique. Thanks for your help, with the second part too. –  Ozzy Apr 28 '13 at 18:15

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