Vectors: Proving a Right Angle with Dot Products & Triangle Area via Cross Product — NYJC 2025 H2 Math Prelim P1
What this question tests
Question
(a) A triangle \(ABC\) is such that \(AB^2 + BC^2 = AC^2\).
By considering \(\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}\) and using the fact that \(|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}\) for any vector \(\mathbf{v}\), prove that \(\angle ABC\) is a right-angle.
(b) In a triangle \(ABC\), the point \(D\) divides \(AC\) in the ratio \(\lambda : 1-\lambda\), where \(0 < \lambda < 1\).
Let the position vectors of \(A\), \(B\), \(C\) and \(D\) be denoted by \(\mathbf{a}\), \(\mathbf{b}\), \(\mathbf{c}\) and \(\mathbf{d}\) respectively.
Show that the area of triangle \(ABD\) is given by \[ k\,|\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}| \] where \(k\) is to be determined in terms of \(\lambda\).
Show full worked solution▾
(a) Proving \(\angle ABC = 90^\circ\)
Square the length of the vector sum, expand using \(|\mathbf{v}|^2 = \mathbf{v}\cdot\mathbf{v}\), then substitute the given Pythagorean relation to force the cross term to vanish.
\[ \begin{aligned} \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC} &\implies \left|\overrightarrow{AB} + \overrightarrow{BC}\right|^2 = \left|\overrightarrow{AC}\right|^2\\ &\implies \left(\overrightarrow{AB}+\overrightarrow{BC}\right)\cdot\left(\overrightarrow{AB}+\overrightarrow{BC}\right) = \overrightarrow{AC}\cdot\overrightarrow{AC} \quad(\because|\mathbf{v}|^2 = \mathbf{v}\cdot\mathbf{v})\\ &\implies \overrightarrow{AB}\cdot\overrightarrow{AB} + 2\left(\overrightarrow{AB}\cdot\overrightarrow{BC}\right) + \overrightarrow{BC}\cdot\overrightarrow{BC} = \overrightarrow{AC}\cdot\overrightarrow{AC}\\ &\implies AB^2 + BC^2 + 2\left(\overrightarrow{AB}\cdot\overrightarrow{BC}\right) = AC^2\\ &\implies AC^2 + 2\left(\overrightarrow{AB}\cdot\overrightarrow{BC}\right) = AC^2 \quad\text{since }AB^2 + BC^2 = AC^2\\ &\implies \overrightarrow{AB}\cdot\overrightarrow{BC} = 0 \end{aligned} \]So \(AB\) is perpendicular to \(BC\) and hence \(\angle ABC = 90^\circ\).
(b) Area of triangle \(ABD\)
Use the ratio theorem to write \(\mathbf{d}\) in terms of \(\mathbf{a}\) and \(\mathbf{c}\), then expand the cross product \(\tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{BD}|\) and collect terms.
By the ratio theorem, \[ \mathbf{d} = \overrightarrow{OD} = (1-\lambda)\overrightarrow{OA} + \lambda\overrightarrow{OC} = (1-\lambda)\mathbf{a} + \lambda\mathbf{c}. \]
Area of triangle \(ABD\): \[ \begin{aligned} &= \tfrac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{BD}\right|\\ &= \tfrac{1}{2}\left|(\mathbf{b}-\mathbf{a})\times(\mathbf{d}-\mathbf{b})\right|\\ &= \tfrac{1}{2}\left|(\mathbf{b}-\mathbf{a})\times\left((1-\lambda)\mathbf{a}+\lambda\mathbf{c}-\mathbf{b}\right)\right|\\ &= \tfrac{1}{2}\left|(1-\lambda)(\mathbf{b}\times\mathbf{a})+\lambda(\mathbf{b}\times\mathbf{c}) -\lambda(\mathbf{a}\times\mathbf{c})+\mathbf{a}\times\mathbf{b}\right|\\ &= \tfrac{1}{2}\left|(\lambda-1)(\mathbf{a}\times\mathbf{b})+\lambda(\mathbf{b}\times\mathbf{c}) -\lambda(\mathbf{a}\times\mathbf{c})\right|\quad\text{since }\mathbf{b}\times\mathbf{a}=-\mathbf{a}\times\mathbf{b}\\ &= \tfrac{1}{2}\left|\lambda(\mathbf{a}\times\mathbf{b})+\lambda(\mathbf{b}\times\mathbf{c}) -\lambda(\mathbf{a}\times\mathbf{c})\right|\\ &= \frac{\lambda}{2}\left|\mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{a}\right| \quad\text{since }0<\lambda<1. \end{aligned} \]
So \(k = \dfrac{\lambda}{2}\).
Alternative: triangles \(ABD\) and \(ABC\) share the same height from \(B\), so their areas are in the ratio \(\lambda : 1\).
\[ \begin{aligned} \text{Area of triangle }ABD &= \frac{\lambda}{\lambda+(1-\lambda)}\left(\text{Area of triangle }ABC\right)\\ &= \lambda\left(\tfrac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|\right)\\ &= \frac{\lambda}{2}\left|(\mathbf{b}-\mathbf{a})\times(\mathbf{c}-\mathbf{a})\right|\\ &= \frac{\lambda}{2}\left|\mathbf{b}\times\mathbf{c}-\mathbf{b}\times\mathbf{a}-\mathbf{a}\times\mathbf{c}\right|\\ &= \frac{\lambda}{2}\left|\mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{a}\right| \end{aligned} \]So \(k = \dfrac{\lambda}{2}\).