3d flash cards

1. 1.Find the shortest distance between two skew lines $\vec r=\vec a_1+\lambda\vec b_1$ and $\vec r=\vec a_2+\mu\vec b_2$.

   $d=\left|\dfrac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|$

   Hint: Use the scalar triple product of the connecting vector with the cross product of direction vectors.

2. 2.What is the perpendicular distance of a point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$?

   $D=\left|\dfrac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

   Hint: Divide the plug-in value by the magnitude of the normal vector.

3. 3.State the condition for two lines with direction cosines $l_1,m_1,n_1$ and $l_2,m_2,n_2$ to be perpendicular, and the relation satisfied by direction cosines of any single line.

   Perpendicular: $l_1l_2+m_1m_2+n_1n_2=0$; also for any line, $l^2+m^2+n^2=1$.

   Hint: Think of direction cosines as components of a unit vector.

4. 4.What is the equation of a plane passing through three non-collinear points $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$, $(x_3,y_3,z_3)$?

   $\begin{vmatrix}x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1\end{vmatrix}=0$

   Hint: Set up a determinant using vectors from one point to the other two.

5. 5.What is the angle $\theta$ between two planes $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$, and what is the condition for them to be perpendicular?

   $\cos\theta=\left|\dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|$; perpendicular when $a_1a_2+b_1b_2+c_1c_2=0$.

   Hint: Compare the angle between their normal vectors.

6. 6.If $l,m,n$ are the direction cosines of a line and $l+m+n=0$, $l^2+m^2+n^2=1$, find the value of $lm+mn+nl$.

   Squaring $l+m+n=0$ gives $l^2+m^2+n^2+2(lm+mn+nl)=0$. Since $l^2+m^2+n^2=1$, we get $1+2(lm+mn+nl)=0$, so $lm+mn+nl=-\dfrac{1}{2}$.

   Hint: Square the linear relation and use $l^2+m^2+n^2=1$.

7. 7.Direction cosines of two lines satisfy $l+m+n=0$ and $2lm+2ln-mn=0$. Find the angle between the two lines.

   Put $l=-(m+n)$ in the second relation: $2(-(m+n))m+2(-(m+n))n-mn=0 \Rightarrow -2m^2-5mn-2n^2=0$, i.e. $2m^2+5mn+2n^2=0=(2m+n)(m+2n)$. So $m=-2n$ (giving direction ratios $(1,-2,1)$) or $n=-2m$ (giving $(1,1,-2)$). Their dot product is $1-2-2=-3$ and each has magnitude $\sqrt6$, so $\cos\theta=\dfrac{-3}{6}=-\dfrac12$, giving the acute angle $\theta=60^\circ$.

   Hint: Eliminate one variable to get a quadratic in the ratio $m/n$.

8. 8.A line makes angles $\alpha,\beta,\gamma,\delta$ with the four diagonals of a cube. Show that $\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta=\dfrac{4}{3}$.

   Take the cube's edges along the axes; the four space diagonals have direction cosines $\dfrac{1}{\sqrt3}(1,1,1)$, $\dfrac{1}{\sqrt3}(1,1,-1)$, $\dfrac{1}{\sqrt3}(1,-1,1)$, $\dfrac{1}{\sqrt3}(-1,1,1)$. If the line has dc's $(l,m,n)$, each $\cos^2$ term is $\dfrac{1}{3}(l\pm m\pm n)^2$; summing all four expands the cross terms to cancel (each of $lm,mn,nl$ appears twice with opposite signs) leaving $\dfrac{1}{3}\cdot4(l^2+m^2+n^2)=\dfrac{4}{3}$, using $l^2+m^2+n^2=1$. Hence the sum equals $\dfrac{4}{3}$.

   Hint: Write each diagonal's dc's as $frac{1}{\sqrt3}(\pm1,\pm1,\pm1)$ and expand.

9. 9.A line has direction ratios $(2,-3,6)$. Find its direction cosines and the angle it makes with the $z$-axis.

   Magnitude $=\sqrt{4+9+36}=\sqrt{49}=7$, so direction cosines are $\left(\dfrac{2}{7},-\dfrac{3}{7},\dfrac{6}{7}\right)$ (or the negatives). The angle with the $z$-axis satisfies $\cos\gamma = n = \dfrac{6}{7}$, so $\gamma=\cos^{-1}\left(\dfrac{6}{7}\right)$.

   Hint: Normalize the direction ratios by their magnitude.

10. 10.The projections of a line segment $AB$ on the coordinate axes are $2,3,6$. Find the length of $AB$ and its direction cosines.

    If $(x_2-x_1,y_2-y_1,z_2-z_1)=(2,3,6)$, the length is $|AB|=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7$. Since projections are the length times direction cosines, $\left(l,m,n\right)=\left(\dfrac{2}{7},\dfrac{3}{7},\dfrac{6}{7}\right)$. So $|AB|=7$ with these direction cosines.

    Hint: Projection on an axis equals length times the corresponding direction cosine.

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