Angle between two lines

Every line in space, however it is written, boils down to a direction — a vector $\vec b$ (or a triple of direction ratios $(a,b,c)$) that tells you which way the line runs. Position along the line does not matter for this question at all; two lines on opposite sides of the room, pointing the same way, are still said to make an angle of $0°$ with each other. Given two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$, the angle $\theta$ between them is defined through the dot product:

where: $\vec{b_1}\cdot\vec{b_2}$ is the dot product of the two direction vectors, $|\vec{b_1}|$ and $|\vec{b_2}|$ are their magnitudes, and the outer modulus forces $\theta$ to come out acute ($0° \le \theta \le 90°$). In Cartesian form, if the lines have direction ratios $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$, the same formula reads:

where: the numerator is the sum of products of matching direction ratios, and each square root under it is the length of that line's direction triple. If direction cosines $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$ are used instead of raw direction ratios, both vectors are already unit vectors, so the formula collapses to the neat $\cos\theta = |l_1l_2+m_1m_2+n_1n_2|$, with no division needed. Two special cases fall straight out of this single formula and are worth memorising as separate facts, since exam questions often ask for them directly instead of the general angle: The lines are parallel exactly when $\theta = 0$, which happens when the direction ratios are proportional, $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$. Geometrically this means the two direction vectors point the same way (or exactly opposite ways), so one is a scalar multiple of the other. The lines are perpendicular exactly when $\theta = 90°$, which happens when the dot product of the direction ratios vanishes, $a_1a_2+b_1b_2+c_1c_2 = 0$. This is the fastest check available: whenever a question only asks you to verify perpendicularity rather than compute a general angle, there is no need to touch the square roots in the denominator at all — just add up the three products and check whether the total is zero. It is worth being clear about what this formula does and does not need. It uses only the direction of each line, never any point that lies on it, so the two position vectors that appear in a vector-form equation of a line (or the numerators $x - x_1$, $y - y_1$, $z - z_1$ in a Cartesian equation) play no role whatsoever in finding the angle — they matter for other three-dimensional geometry questions, such as finding where two lines meet or how far apart two skew lines are, but not for this one.

Problem. Find the angle between the lines $\vec r = (2\hat\imath - 5\hat\jmath + \hat k) + \lambda(3\hat\imath+2\hat\jmath+6\hat k)$ and $\vec r = (7\hat\imath - 6\hat k) + \mu(\hat\imath+2\hat\jmath+2\hat k)$.

Step 1. Read off the direction vectors, ignoring the position vectors entirely: $\vec{b_1} = 3\hat\imath+2\hat\jmath+6\hat k$ and $\vec{b_2} = \hat\imath+2\hat\jmath+2\hat k$.

Step 2. Compute the dot product: $\vec{b_1}\cdot\vec{b_2} = (3)(1)+(2)(2)+(6)(2) = 3+4+12 = 19$.

Step 3. Compute each magnitude: $|\vec{b_1}| = \sqrt{3^2+2^2+6^2} = \sqrt{49} = 7$ and $|\vec{b_2}| = \sqrt{1^2+2^2+2^2} = \sqrt{9} = 3$.

Step 4. Substitute into the formula: $\cos\theta = \left|\dfrac{19}{7\times 3}\right| = \dfrac{19}{21}$, so $\theta = \cos^{-1}\left(\dfrac{19}{21}\right)$.

Competitive stretch

Problem. Show that $\dfrac{x-5}{7} = \dfrac{y+2}{-5} = \dfrac{z}{1}$ and $\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}$ are perpendicular.

Step 1. Read off direction ratios: line 1 is $(7,-5,1)$, line 2 is $(1,2,3)$.

Step 2. Compute $a_1a_2+b_1b_2+c_1c_2 = (7)(1)+(-5)(2)+(1)(3) = 7-10+3 = 0$.

Step 3. Since the dot product of the direction ratios is $0$, $\cos\theta = 0$, so $\theta = 90°$ and the lines are perpendicular — no need to ever compute the full angle when only perpendicularity is asked. Contrast the two examples: the first genuinely asks for an angle, so all four steps — dot product, both magnitudes, division, inverse cosine — are needed and the answer is left as $\cos^{-1}(19/21)$ rather than forced into a decimal. The second only asks for a yes/no verification, so the magnitudes are never even computed; recognising which kind of question you are facing saves real time in an exam.

• Dropping the modulus. Without it, a negative numerator would suggest an obtuse angle, but the angle between two lines is conventionally reported as acute.
• Treating proportional direction ratios as "the same line." Parallel direction only tells you about orientation, not location — the lines can still be distinct and non-intersecting.
• Reading direction ratios off a non-standard Cartesian form, e.g. $2x=3y=6z$, without first converting to symmetric form $\dfrac{x}{1/2}=\dfrac{y}{1/3}=\dfrac{z}{1/6}$.
• Sign slips, especially with a negative direction ratio like $\dfrac{y+2}{-5}$ giving $b_1=-5$ — dropping the sign wrecks the whole dot product.

Ask the tutor: Want to check whether two lines you have are parallel, perpendicular, or at some other angle — try it with the AI tutor.