Easy Practice — Set 1

Quick Recap — Points in Space

• Distance: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$; from the origin, $\sqrt{x^2+y^2+z^2}$.
• Midpoint / section formula extend from 2D by adding the $z$-term.
• Space is divided into 8 octants; the coordinate planes are $x=0$ (yz), $y=0$ (xz), $z=0$ (xy).
• Direction cosines $l,m,n$ satisfy $l^2+m^2+n^2=1$.

Beyond-NCERT JEE Formulae

Everything below is powered by the scalar triple product $[\vec a\ \vec b\ \vec c]=\vec a\cdot(\vec b\times\vec c)$, the single most useful object in JEE 3D geometry.

1. Shortest distance between two skew lines. For $\vec r=\vec a_1+\lambda\vec b_1$ and $\vec r=\vec a_2+\mu\vec b_2$, $$d=\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}.$$ [JEE Tip] The numerator is the scalar triple product $[\,\vec a_2-\vec a_1\ \ \vec b_1\ \ \vec b_2\,]$. If it comes out $0$ the lines are not skew — they are coplanar (intersecting or parallel), and the shortest distance collapses to the point-to-line case.

2. Coplanarity / intersection of two lines. The two lines above are coplanar (and, when $\vec b_1$ is not parallel to $\vec b_2$, they actually intersect) iff $$(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=0,$$ equivalently the $3\times3$ determinant with rows $\vec a_2-\vec a_1$, $\vec b_1$, $\vec b_2$ vanishes: $$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}=0.$$ This determinant is the fastest tool for "find the parameter that makes the two lines meet" problems.

3. Angle between a line and a plane. For a line of direction $\vec b$ and a plane of normal $\vec n$, $$\sin\theta=\dfrac{|\vec b\cdot\vec n|}{|\vec b|\,|\vec n|}.$$ [JEE Tip] Note the sine, not cosine: $\theta$ is the complement of the line-normal angle. Hence $\vec b\cdot\vec n=0$ means the line is parallel to (or lies in) the plane, while $\vec b$ parallel to $\vec n$ means it is perpendicular to the plane.

4. Distance between two parallel lines. For $\vec r=\vec a_1+\lambda\vec b$ and $\vec r=\vec a_2+\mu\vec b$ (same direction $\vec b$), $$d=\dfrac{|(\vec a_2-\vec a_1)\times\vec b|}{|\vec b|}.$$ This is just the length of the part of $\vec a_2-\vec a_1$ perpendicular to $\vec b$ — the cross product removes the parallel component.

5. Volume of a tetrahedron. With coterminous edge vectors $\vec a,\vec b,\vec c$ meeting at one vertex, $$V=\dfrac{1}{6}\,\big|[\vec a\ \ \vec b\ \ \vec c]\big|=\dfrac{1}{6}\,|\vec a\cdot(\vec b\times\vec c)|.$$ From four vertices $A,B,C,D$ take $\vec a=\overrightarrow{AB}$, $\vec b=\overrightarrow{AC}$, $\vec c=\overrightarrow{AD}$. A value of $0$ means the four points are coplanar.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Shortest distance between skew lines (Cartesian). Find the shortest distance between $$L_1:\ \dfrac{x-3}{1}=\dfrac{y-5}{-2}=\dfrac{z-7}{1},\qquad L_2:\ \dfrac{x+1}{7}=\dfrac{y+1}{-6}=\dfrac{z+1}{1}.$$ Read off $\vec a_1=(3,5,7)$, $\vec b_1=(1,-2,1)$ and $\vec a_2=(-1,-1,-1)$, $\vec b_2=(7,-6,1)$, so $\vec a_2-\vec a_1=(-4,-6,-8)$. The cross product is $\vec b_1\times\vec b_2=(4,6,8)$ with $|\vec b_1\times\vec b_2|=\sqrt{16+36+64}=\sqrt{116}=2\sqrt{29}\neq0$, so the lines are skew. The triple product is $(-4)(4)+(-6)(6)+(-8)(8)=-116$, hence $$d=\dfrac{|-116|}{2\sqrt{29}}=\dfrac{58}{\sqrt{29}}=2\sqrt{29}.$$

Example 2 — Shortest distance between skew lines (vector form). Take $\vec r=(\hat i+\hat j)+\lambda(2\hat i-\hat j+\hat k)$ and $\vec r=(2\hat i+\hat j-\hat k)+\mu(3\hat i-5\hat j+2\hat k)$. Here $\vec a_2-\vec a_1=(1,0,-1)$, $\vec b_1=(2,-1,1)$ and $\vec b_2=(3,-5,2)$. Then $\vec b_1\times\vec b_2=(3,-1,-7)$ with $|\vec b_1\times\vec b_2|=\sqrt{9+1+49}=\sqrt{59}$, and the triple product is $(1)(3)+(0)(-1)+(-1)(-7)=10$. Therefore $$d=\dfrac{10}{\sqrt{59}}\approx1.30.$$

Example 3 — Coplanarity test and point of intersection. Show that $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x-4}{5}=\dfrac{y-1}{2}=\dfrac{z}{1}$ are coplanar, and find where they meet. With $\vec a_2-\vec a_1=(3,-1,-3)$, $\vec b_1=(2,3,4)$ and $\vec b_2=(5,2,1)$, $$\begin{vmatrix} 3 & -1 & -3 \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix}=3(3-8)+1(2-20)-3(4-15)=-15-18+33=0,$$ so the lines are coplanar. Since $\vec b_1$ and $\vec b_2$ are not proportional the lines are not parallel, so they intersect. Writing them as $(1+2t,2+3t,3+4t)$ and $(4+5s,1+2s,s)$, the $z$- and $y$-equations give $t=-1$ and $s=-1$ (the $x$-equation checks). Substituting $t=-1$ gives the point of intersection $(-1,-1,-1)$.

Example 4 — Angle between a line and a plane. Find the acute angle between the line $\dfrac{x-1}{1}=\dfrac{y-2}{1}=\dfrac{z-3}{2}$ and the plane $2x-y+z=4$. The direction is $\vec b=(1,1,2)$ and the normal is $\vec n=(2,-1,1)$, so $\vec b\cdot\vec n=2-1+2=3$ with $|\vec b|=\sqrt6$ and $|\vec n|=\sqrt6$. Using the sine form, $$\sin\theta=\dfrac{|3|}{\sqrt6\cdot\sqrt6}=\dfrac{3}{6}=\dfrac12\ \Rightarrow\ \theta=30^\circ.$$ [JEE Tip] applying a cosine here would wrongly give $60^\circ$.

Example 5 — Volume of a tetrahedron from four vertices. For $A(1,1,1)$, $B(2,3,4)$, $C(4,2,3)$ and $D(3,4,2)$, the coterminous edges are $\overrightarrow{AB}=(1,2,3)$, $\overrightarrow{AC}=(3,1,2)$ and $\overrightarrow{AD}=(2,3,1)$. Their scalar triple product is $$\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix}=1(1-6)-2(3-4)+3(9-2)=-5+2+21=18,$$ so $V=\dfrac16|18|=3$ cubic units. (A value of $0$ would have meant the four vertices were coplanar.)

Example 6 — Distance between two parallel lines. Take $\vec r=(\hat i+2\hat j-3\hat k)+\lambda(2\hat i-\hat j+2\hat k)$ and $\vec r=(3\hat i-\hat j+\hat k)+\mu(2\hat i-\hat j+2\hat k)$; the common direction is $\vec b=(2,-1,2)$ with $|\vec b|=3$. With $\vec a_2-\vec a_1=(2,-3,4)$, the cross product is $(\vec a_2-\vec a_1)\times\vec b=(-2,4,4)$ of magnitude $\sqrt{4+16+16}=6$, so $$d=\dfrac{6}{3}=2.$$

Example 7 — Parameter value for coplanarity. Find $\lambda$ so that $\dfrac{x-1}{1}=\dfrac{y+1}{-1}=\dfrac{z}{2}$ and $\dfrac{x-2}{2}=\dfrac{y-1}{1}=\dfrac{z-1}{\lambda}$ are coplanar. Here $\vec a_2-\vec a_1=(1,2,1)$, $\vec b_1=(1,-1,2)$ and $\vec b_2=(2,1,\lambda)$, and coplanarity requires $$\begin{vmatrix} 1 & 2 & 1 \\ 1 & -1 & 2 \\ 2 & 1 & \lambda \end{vmatrix}=1(-\lambda-2)-2(\lambda-4)+1(1+2)=-3\lambda+9=0,$$ so $\lambda=3$.