Line and Plane — MHT-CET Mathematics

Formulas (8)

• Direction ratios, direction cosines, and the l² + m² + n² = 1 identity · Direction cosines and their identity
  $$(l, m, n) = \frac{(a, b, c)}{\sqrt{a^2 + b^2 + c^2}}, \qquad l^2 + m^2 + n^2 = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$
• Symmetric Cartesian form and vector form of a line · Symmetric and vector form
  $$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \quad\Longleftrightarrow\quad \vec{r} = \vec{a} + \lambda\vec{b}$$
• Line through two points · Direction from two points
  $$\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$$
• Normalising a non-standard Cartesian equation · Reduce to symmetric form
  $$px - p_0 = qy - q_0 = rz - r_0 \;\Rightarrow\; \frac{x - p_0/p}{1/p} = \frac{y - q_0/q}{1/q} = \frac{z - r_0/r}{1/r}$$
• Direction as a cross product: ⊥ two lines, ∥ two planes, intersection of two planes · Cross product (determinant expansion)
  $$\vec{u} \times \vec{v} = (u_2 v_3 - u_3 v_2)\,\hat{i} - (u_1 v_3 - u_3 v_1)\,\hat{j} + (u_1 v_2 - u_2 v_1)\,\hat{k}$$
• Unit vector perpendicular to two lines · Unit normal to two lines
  $$\hat{n} = \frac{\vec{d}_1 \times \vec{d}_2}{|\vec{d}_1 \times \vec{d}_2|}$$
• Angle a line makes with an axis; equal inclination · Angle with an axis
  $$\cos\alpha = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \qquad \text{equally inclined} \Rightarrow |a| = |b| = |c|,\ \cos = \tfrac{1}{\sqrt 3}$$
• Direction cosines from a linear + quadratic constraint pair · Method (linear eliminate, quadratic factor, normalise)
  $$l = 5m - 3n \;\xrightarrow{\text{sub}}\; \text{quadratic in } m, n \;\Rightarrow\; l : m : n \;\xrightarrow{\div\sqrt{l^2+m^2+n^2}}\; (l, m, n)$$

Watch out for (17)

• Direction ratios are NOT direction cosines until you normalise
  → Direction ratios, direction cosines, and the l² + m² + n² = 1 identity
• The $\pm$ sign decides acute vs obtuse
  → Direction ratios, direction cosines, and the l² + m² + n² = 1 identity
• Numerators give the point, denominators give the direction — don't swap them
  → Symmetric Cartesian form and vector form of a line
• A fixed coordinate means a zero direction component
  → Symmetric Cartesian form and vector form of a line
• Head minus tail — keep the order consistent
  → Line through two points
• "Parallel to the line joining P, Q" ≠ "through P or Q"
  → Line through two points
• You must factor BEFORE reading the point
  → Normalising a non-standard Cartesian equation
• Direction ratios are the RECIPROCALS of the coefficients
  → Normalising a non-standard Cartesian equation
• Parallel to two PLANES → cross the NORMALS, not the planes
  → Direction as a cross product: ⊥ two lines, ∥ two planes, intersection of two planes
• The $\hat{j}$ component flips sign
  → Direction as a cross product: ⊥ two lines, ∥ two planes, intersection of two planes
• Direction ratios are only defined up to a scalar
  → Direction as a cross product: ⊥ two lines, ∥ two planes, intersection of two planes
• Factor the cross product before normalising
  → Unit vector perpendicular to two lines
• Both signs are valid — read the option set
  → Unit vector perpendicular to two lines
• Divide by the magnitude — $\cos\alpha = a/|\vec{d}|$, not $a$ alone
  → Angle a line makes with an axis; equal inclination
• "Equally inclined" means equal magnitudes, watch the signs
  → Angle a line makes with an axis; equal inclination
• Solve for the RATIO first, then normalise — don't skip normalisation
  → Direction cosines from a linear + quadratic constraint pair
• Two roots → two answers; report both if asked
  → Direction cosines from a linear + quadratic constraint pair

Formulas (12)

• Equation of a plane and its normal · The three equivalent forms
  $$ax + by + cz = d \;\Longleftrightarrow\; \vec{r}\cdot\vec{n} = d \;\Longleftrightarrow\; \vec{n}\cdot(\vec{r} - \vec{a}) = 0$$
• Direction cosines of the normal · Direction-cosine identity
  $$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$
• Planes parallel to a coordinate plane or to a given plane · Same normal, new constant
  $$ax + by + cz = d' \quad\text{where } d' = a x_0 + b y_0 + c z_0$$
• Plane from the foot of the perpendicular from the origin · Plane from foot of perpendicular
  $$\vec{r}\cdot\overrightarrow{OM} = |\overrightarrow{OM}|^2 = x_0^2 + y_0^2 + z_0^2$$
• Plane through a point with normal fixed by axis angles · Point-normal with angle-derived normal
  $$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$
• Plane perpendicular to two given planes · Normal from two perpendicular planes
  $$\vec{n} = \vec{n_1}\times\vec{n_2}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0$$
• Plane through a point parallel to two lines · Normal from two parallel lines
  $$\vec{n} = \vec{d_1}\times\vec{d_2}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0$$
• Plane through three points · Three-point plane
  $$\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}, \qquad \vec{n}\cdot(\vec{r} - \vec{a}) = 0$$
• Perpendicular bisector plane of a segment · Perpendicular bisector plane
  $$\vec{n} = \overrightarrow{PQ}, \quad M = \tfrac{1}{2}(\vec{p} + \vec{q}), \quad \vec{n}\cdot(\vec{r} - \vec{m}) = 0$$
• Family of planes through a line of intersection (lambda engine) · Family of planes
  $$P_1 + \lambda P_2 = 0, \qquad \vec{n}_\lambda = \big(a_1 + \lambda a_2,\; b_1 + \lambda b_2,\; c_1 + \lambda c_2\big)$$
• Intercept form, intercept triangle area and centroid · Intercept triangle: centroid and area
  $$G = \left(\tfrac{a}{3}, \tfrac{b}{3}, \tfrac{c}{3}\right), \qquad \text{Area} = \tfrac{1}{2}\sqrt{a^2 b^2 + b^2 c^2 + c^2 a^2}$$
• Recovering a plane from a point and its mirror image · Plane from point and its image
  $$M = \tfrac12(P + P'), \quad \vec{n} = P' - P, \quad \vec{n}\cdot(\vec{r} - \vec{m}) = 0$$

Watch out for (25)

• The normal is the coefficient triple, not the point
  → Equation of a plane and its normal
• $d$ is found by substituting, never left at the wrong sign
  → Equation of a plane and its normal
• "Acute angle" chooses the positive square root
  → Direction cosines of the normal
• Equally inclined means equal COSINES, not equal angles spread over 90 degrees
  → Direction cosines of the normal
• Parallel to XY-plane is $z = k$, not $x + y = k$
  → Planes parallel to a coordinate plane or to a given plane
• Re-use the WHOLE normal when copying a plane
  → Planes parallel to a coordinate plane or to a given plane
• The constant is $|\overrightarrow{OM}|^2$, not $|\overrightarrow{OM}|$
  → Plane from the foot of the perpendicular from the origin
• Don't move the foot to the wrong side of the equation
  → Plane from the foot of the perpendicular from the origin
• Clear the irrational direction cosine into a clean ratio
  → Plane through a point with normal fixed by axis angles
• Put the point into the expanded form, not the angle data
  → Plane through a point with normal fixed by axis angles
• Cross product, not dot product, for the normal
  → Plane perpendicular to two given planes
• Keep the cross-product sign and middle-term flip straight
  → Plane perpendicular to two given planes
• Parallel to two LINES uses their directions, parallel to two PLANES uses their normals
  → Plane through a point parallel to two lines
• Read line directions from the denominators, signs included
  → Plane through a point parallel to two lines
• Anchor BOTH edge vectors at the same point
  → Plane through three points
• "Parallel to an axis" kills exactly one coefficient
  → Plane through three points
• Pass through the MIDPOINT, not through $P$ or $Q$
  → Perpendicular bisector plane of a segment
• Simplify the normal before substituting
  → Perpendicular bisector plane of a segment
• Perpendicular to the XY-plane means the Z-coefficient vanishes
  → Family of planes through a line of intersection (lambda engine)
• Parallel-to-axis kills the SAME-named coefficient
  → Family of planes through a line of intersection (lambda engine)
• Clear the fractions before matching options
  → Family of planes through a line of intersection (lambda engine)
• Read intercepts from the form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
  → Intercept form, intercept triangle area and centroid
• Use the squared intercepts in the area formula
  → Intercept form, intercept triangle area and centroid
• The normal is the segment, the plane is at the midpoint
  → Recovering a plane from a point and its mirror image
• Simplify the messy normal before testing option-points
  → Recovering a plane from a point and its mirror image

Formulas (8)

• Direction ratios, direction cosines, and the dot/cross toolkit · The toolkit
  $$l^2 + m^2 + n^2 = 1, \qquad \vec{u}\cdot\vec{v} = u_1v_1 + u_2v_2 + u_3v_3, \qquad |\vec{d}| = \sqrt{a^2 + b^2 + c^2}$$
• Angle between two lines · Angle between two lines
  $$\cos\theta = \frac{|\vec{d_1}\cdot\vec{d_2}|}{|\vec{d_1}|\,|\vec{d_2}|}$$
• Angle between two planes (and solving for an unknown coefficient) · Angle between two planes
  $$\cos\theta = \frac{|\vec{n_1}\cdot\vec{n_2}|}{|\vec{n_1}|\,|\vec{n_2}|}, \qquad |\alpha_1 - \alpha_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}$$
• Angle between a line and a plane (the solve-for-lambda variant) · Angle between a line and a plane
  $$\sin\theta = \frac{|\vec{d}\cdot\vec{n}|}{|\vec{d}|\,|\vec{n}|}$$
• Parallel and perpendicular conditions (lines, and line-parallel-to-plane) · Perpendicular / parallel conditions
  $$\vec{d_1}\cdot\vec{d_2} = 0 \;\text{(}\perp\text{)}, \qquad \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \;\text{(}\parallel\text{)}, \qquad \vec{d}\cdot\vec{n} = 0 \;\text{(line}\parallel\text{plane)}$$
• Line lies in a plane (two conditions, solve the unknowns) · Line-lies-in-plane conditions
  $$Ax_1 + By_1 + Cz_1 = d \quad \text{and} \quad aA + bB + cC = 0$$
• Direction-cosine systems and equal-angle lines · Direction-cosine identity
  $$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1, \qquad \cos\theta = |l_1l_2 + m_1m_2 + n_1n_2|$$
• Line of intersection of two planes, and its angle with an axis · Line of intersection
  $$\vec{d} = \vec{n_1}\times\vec{n_2}, \qquad \cos\alpha = \frac{|\vec{d}\cdot\hat{x}|}{|\vec{d}|}$$

Watch out for (16)

• Direction RATIOS are not direction COSINES
  → Direction ratios, direction cosines, and the dot/cross toolkit
• A zero denominator is a valid direction ratio
  → Direction ratios, direction cosines, and the dot/cross toolkit
• Drop the modulus and you may report the obtuse angle
  → Angle between two lines
• Lines need DIRECTIONS, not points
  → Angle between two lines
• Plane angle uses normals, not the plane's 'direction'
  → Angle between two planes (and solving for an unknown coefficient)
• The question may want the DIFFERENCE of roots, not a root
  → Angle between two planes (and solving for an unknown coefficient)
• Use SINE for line–plane, COSINE for line–line and plane–plane
  → Angle between a line and a plane (the solve-for-lambda variant)
• Convert a $\cos^{-1}$ given-angle to $\sin\theta$ first
  → Angle between a line and a plane (the solve-for-lambda variant)
• Line PARALLEL to a plane means direction ⟂ NORMAL
  → Parallel and perpendicular conditions (lines, and line-parallel-to-plane)
• Normalise messy ratios before dotting
  → Parallel and perpendicular conditions (lines, and line-parallel-to-plane)
• BOTH conditions are required — one is not enough
  → Line lies in a plane (two conditions, solve the unknowns)
• Read the point off the numerators correctly
  → Line lies in a plane (two conditions, solve the unknowns)
• Use the identity $\sum\cos^2 = 1$, not $\sum\cos = 1$
  → Direction-cosine systems and equal-angle lines
• A two-constraint system gives TWO directions — find the angle BETWEEN them
  → Direction-cosine systems and equal-angle lines
• The intersection direction is the CROSS product of the normals
  → Line of intersection of two planes, and its angle with an axis
• Plane 'parallel to two vectors' ⟹ normal is THEIR cross product
  → Line of intersection of two planes, and its angle with an axis

Formulas (8)

• Distance of a point from the axes and the origin · Distance from origin and axes
  $$OP = \sqrt{x^2+y^2+z^2}, \qquad \sum(\text{axis distance})^2 = 2(x^2+y^2+z^2)$$
• Distance of a point from a plane · Point-to-plane distance
  $$d = \frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$
• Equidistant points and the gap between parallel planes · Distance between parallel planes
  $$d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$$
• Distance of a point from a line · Point-to-line distance
  $$d = \frac{|\overrightarrow{AP} \times \vec{b}|}{|\vec{b}|} = \sqrt{|\overrightarrow{AP}|^2 - \left(\frac{\overrightarrow{AP}\cdot\vec{b}}{|\vec{b}|}\right)^2}$$
• Distance between two parallel lines · Distance between parallel lines
  $$d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$$
• Shortest distance between skew lines (and solving backwards for a parameter) · Shortest distance between skew lines
  $$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$
• Build a plane from conditions, then take a distance · Plane normal from a cross product
  $$\vec{n} = \vec{p} \times \vec{q}, \qquad d = \frac{|\vec{n}\cdot(\vec{r}_1 - \vec{r}_0)|}{|\vec{n}|}$$
• Where a line meets a plane, and distance measured along a line · Line in parametric form
  $$(x, y, z) = (x_0 + at,\ y_0 + bt,\ z_0 + ct)$$

Watch out for (16)

• Axis distance DROPS one coordinate, origin distance keeps all three
  → Distance of a point from the axes and the origin
• Sum-of-squares from axes is TWICE the origin-squared, not equal
  → Distance of a point from the axes and the origin
• Move every term to one side first — the constant $d$ must be in $=0$ form
  → Distance of a point from a plane
• Absolute value on top — distance is never negative
  → Distance of a point from a plane
• Equidistant gives TWO cases — keep both signs
  → Equidistant points and the gap between parallel planes
• Parallel-plane gap needs MATCHING normals
  → Equidistant points and the gap between parallel planes
• Divide by $|\vec{b}|$, not by $|\vec{b}|^2$
  → Distance of a point from a line
• $\overrightarrow{AP} = P - A$ — point minus the line's point
  → Distance of a point from a line
• Use the JOIN vector $\vec{a}_2 - \vec{a}_1$, not a single point
  → Distance between two parallel lines
• Confirm parallel FIRST
  → Distance between two parallel lines
• Numerator is a scalar (dot of difference with the cross), denominator is the cross's MAGNITUDE
  → Shortest distance between skew lines (and solving backwards for a parameter)
• Backwards problems often hide TWO roots — pick by the stated constraint
  → Shortest distance between skew lines (and solving backwards for a parameter)
• The normal is the CROSS product, then the plane passes through the GIVEN point
  → Build a plane from conditions, then take a distance
• Simplify the normal before plugging in
  → Build a plane from conditions, then take a distance
• 'Measured along the line' ≠ perpendicular distance
  → Where a line meets a plane, and distance measured along a line
• Equal angles with the axes fixes the direction to $(1,1,1)$
  → Where a line meets a plane, and distance measured along a line

Formulas (7)

• Foot of the perpendicular from a point to a line · Foot on a line
  $$F = \vec{a} + \lambda\vec{d}, \qquad \overrightarrow{PF}\cdot\vec{d} = 0 \;\Rightarrow\; \lambda$$
• Foot of the perpendicular from a point to a plane · Foot on a plane
  $$F = P + t\,\vec{n}, \qquad t = -\dfrac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$$
• Mirror image of a point in a plane · Image in a plane
  $$P' = 2F - P = P + 2t\,\vec{n}, \qquad t = -\dfrac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$$
• Mirror image of a point in a line · Image in a line (forward and backward)
  $$P' = 2F - P; \qquad \text{midpoint } \tfrac{P+P'}{2} \in \text{line}, \;\; \overrightarrow{PP'}\cdot\vec{d} = 0$$
• Image of a line in a plane (and planes through an image) · Image line — reflect point, preserve direction
  $$\vec{d}\cdot\vec{n} = 0 \;\Rightarrow\; \text{line} \parallel \text{plane}, \quad \text{image line} = \{P',\; \vec{d}\}$$
• Projection of a segment onto a line · Projection onto a line
  $$\text{proj} = \dfrac{|\overrightarrow{AB}\cdot\vec{d}|}{|\vec{d}|}$$
• Projection of a segment onto a plane · Projection onto a plane
  $$\text{proj}_{\text{plane}} = \sqrt{|\overrightarrow{AB}|^2 - \left(\dfrac{\overrightarrow{AB}\cdot\vec{n}}{|\vec{n}|}\right)^2}$$

Watch out for (16)

• Perpendicularity is $\overrightarrow{PF}\cdot\vec{d}=0$, NOT $\overrightarrow{PF} = \vec{d}$
  → Foot of the perpendicular from a point to a line
• Use the symmetric form's parameter consistently
  → Foot of the perpendicular from a point to a line
• Don't forget to substitute back
  → Foot of the perpendicular from a point to a line
• Carry the constant $d$ with its correct sign
  → Foot of the perpendicular from a point to a plane
• Divide by $a^2+b^2+c^2$, not by $\sqrt{a^2+b^2+c^2}$
  → Foot of the perpendicular from a point to a plane
• Image is $2F - P$, the foot is only halfway
  → Mirror image of a point in a plane
• $2F - P$, not $F - 2P$ or $2P - F$
  → Mirror image of a point in a plane
• Backward problems use TWO conditions, not one
  → Mirror image of a point in a line
• Reflect in the LINE, not the line's fixed point
  → Mirror image of a point in a line
• Preserve the direction ratios — don't negate them
  → Image of a line in a plane (and planes through an image)
• Reflect a point ON the line, then reattach the direction
  → Image of a line in a plane (and planes through an image)
• Divide by $|\vec{d}|$, not $|\vec{d}|^2$
  → Projection of a segment onto a line
• $\overrightarrow{AB} = B - A$ — direction matters inside the dot product
  → Projection of a segment onto a line
• Plane projection SUBTRACTS the normal part; line projection KEEPS the direction part
  → Projection of a segment onto a plane
• Use the UNIT normal inside the square
  → Projection of a segment onto a plane
• Answer is $\sqrt{2/3}$, not $2/3$
  → Projection of a segment onto a plane

Formulas (9)

• A general point on a line · General point on a line
  $$P = (x_1 + at,\; y_1 + bt,\; z_1 + ct) \qquad \vec{r} = \vec{a} + t\,\vec{d}$$
• Point where a line meets a plane · Line meets plane
  $$A(x_1+at)+B(y_1+bt)+C(z_1+ct)=D \;\Rightarrow\; t \;\Rightarrow\; P$$
• Point of intersection of two lines · Intersection by equating
  $$x_1+a_1t = x_2+a_2s,\quad y_1+b_1t = y_2+b_2s,\quad z_1+c_1t = z_2+c_2s$$
• Coplanarity and intersect-find-k by the scalar triple product · Coplanarity / intersection determinant = 0
  $$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}=0$$
• Four points coplanar · Four points coplanar
  $$[\,\vec{AB},\;\vec{AC},\;\vec{AD}\,]=0$$
• Shortest distance between skew lines · Shortest distance (skew lines)
  $$d = \frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|}$$
• Direction of the line of intersection of two planes · Direction of line of intersection of two planes
  $$\vec{d}=\vec{n_1}\times\vec{n_2}$$
• Transversal intersecting two given lines · Transversal condition
  $$\vec{AB}\parallel (l,m,n) \;\Rightarrow\; \frac{(\vec{AB})_x}{l}=\frac{(\vec{AB})_y}{m}=\frac{(\vec{AB})_z}{n}$$
• Condition for a line to lie in a plane · Line lies in plane (both conditions)
  $$\vec{d}\cdot\vec{n}=0 \quad\text{and}\quad P_0 \in \text{plane}$$

Watch out for (19)

• A NEGATIVE denominator is a negative direction ratio
  → A general point on a line
• Use DIFFERENT parameters for two different lines
  → A general point on a line
• XZ-plane is $y = 0$, not $z = 0$
  → Point where a line meets a plane
• The question may want a derived quantity, not the point itself
  → Point where a line meets a plane
• Always verify the THIRD equation
  → Point of intersection of two lines
• Read the FINAL ask
  → Point of intersection of two lines
• The QUADRATIC trap — there are usually TWO values of k
  → Coplanarity and intersect-find-k by the scalar triple product
• Joining vector is $A_2 - A_1$, and it is ROW 1
  → Coplanarity and intersect-find-k by the scalar triple product
• 'Intersect' uses the SAME determinant as 'coplanar'
  → Coplanarity and intersect-find-k by the scalar triple product
• All edge vectors must start from the SAME base point
  → Four points coplanar
• Four-point coplanarity is usually LINEAR in the unknown
  → Four points coplanar
• Divide by $|\vec{d_1}\times\vec{d_2}|$, and take the ABSOLUTE value on top
  → Shortest distance between skew lines
• 'SD given, find the parameter' is a QUADRATIC — expect two values
  → Shortest distance between skew lines
• Use the NORMALS, not the planes' constants
  → Direction of the line of intersection of two planes
• Cross-product sign — keep the middle term's minus
  → Direction of the line of intersection of two planes
• VERIFY the parallel condition after solving
  → Transversal intersecting two given lines
• Two SEPARATE parameters, one per line
  → Transversal intersecting two given lines
• BOTH conditions are required
  → Condition for a line to lie in a plane
• Perpendicular DIRECTIONS, parallel LINE
  → Condition for a line to lie in a plane

Formulas (3)

• Centroid of a tetrahedron and a triangle · Centroid (tetrahedron and triangle)
  $$G_{\text{tet}} = \dfrac{A + B + C + D}{4} \qquad G_{\triangle} = \dfrac{A + B + C}{3}$$
• Volume of a tetrahedron via the scalar triple product · Tetrahedron volume
  $$V = \dfrac{1}{6}\left|\det\begin{bmatrix} \overrightarrow{AB} \\ \overrightarrow{AC} \\ \overrightarrow{AD} \end{bmatrix}\right| = \dfrac{1}{6}\big|\,\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD})\,\big|$$
• Volume of OABC from a plane cutting the axes · Plane normal, intercepts, and OABC volume
  $$\vec{n} = \vec{d_1}\times\vec{d_2} \qquad V_{OABC} = \dfrac{1}{6}\,|a\,b\,c| = \dfrac{k^3}{6}\ \ (\text{for } x+y+z=k)$$

Watch out for (9)

• Divide by 4 for a tetrahedron, by 3 for a triangle
  → Centroid of a tetrahedron and a triangle
• Inverse problems: rearrange, don't re-guess
  → Centroid of a tetrahedron and a triangle
• Watch which coordinate the puzzle reuses
  → Centroid of a tetrahedron and a triangle
• It's $\tfrac{1}{6}$ for a tetrahedron, not $\tfrac{1}{3}$ or 1
  → Volume of a tetrahedron via the scalar triple product
• Build edges from ONE common vertex
  → Volume of a tetrahedron via the scalar triple product
• Set $|\det| = 6V$, not $|\det| = V$
  → Volume of a tetrahedron via the scalar triple product
• Normal = cross product, then the constant comes from the POINT
  → Volume of OABC from a plane cutting the axes
• Volume of OABC is $\tfrac{1}{6}\,abc$, not $\tfrac{1}{6}k$ or $abc$
  → Volume of OABC from a plane cutting the axes
• Simplify the cross-product normal before reading intercepts
  → Volume of OABC from a plane cutting the axes