IB Mathematics SL 3.1 The distance between two points AI SL Paper 2 - Exam Style Questions - New Syllabus
▶️ Answer/Explanation
Markscheme-style working
(a) Distance \(BV\)
Method A (3D distance formula)
\[ BV=\sqrt{(8-4)^2+(6-3)^2+(0-10)^2} =\sqrt{16+9+100} =\sqrt{125}=5\sqrt{5}\approx \boxed{11.2}. \] Method B (two-step Pythagoras)
Horizontal gap \(=\sqrt{(8-4)^2+(6-3)^2}=5\). Then \(BV=\sqrt{5^2+10^2}=\sqrt{125}=5\sqrt{5}\).
\[ BV=\sqrt{(8-4)^2+(6-3)^2+(0-10)^2} =\sqrt{16+9+100} =\sqrt{125}=5\sqrt{5}\approx \boxed{11.2}. \] Method B (two-step Pythagoras)
Horizontal gap \(=\sqrt{(8-4)^2+(6-3)^2}=5\). Then \(BV=\sqrt{5^2+10^2}=\sqrt{125}=5\sqrt{5}\).
(b) Angle \(\widehat{BVC}\)
First note \(VB=VC=\sqrt{(4-8)^2+(3-6)^2+(10-0)^2}=\sqrt{125}\) and \(BC=\sqrt{(8-8)^2+(6-0)^2+(0-0)^2}=6\).
Method 1 (Cosine rule in \(\triangle BVC\))
\[ \cos\angle BVC =\frac{VB^2+VC^2-BC^2}{2\cdot VB\cdot VC} =\frac{125+125-36}{2\cdot\sqrt{125}\cdot\sqrt{125}} =\frac{214}{250}=\frac{107}{125}. \] \[ \angle BVC=\cos^{-1}\!\left(\frac{107}{125}\right) \approx \boxed{0.543\ \text{rad}}\ \ (\approx \boxed{31.1^\circ}). \]
\[ \cos\angle BVC =\frac{VB^2+VC^2-BC^2}{2\cdot VB\cdot VC} =\frac{125+125-36}{2\cdot\sqrt{125}\cdot\sqrt{125}} =\frac{214}{250}=\frac{107}{125}. \] \[ \angle BVC=\cos^{-1}\!\left(\frac{107}{125}\right) \approx \boxed{0.543\ \text{rad}}\ \ (\approx \boxed{31.1^\circ}). \]
Method 2 (Midpoint/right-triangle approach)
Let \(M\) be the midpoint of \(BC\) so \(M(8,3,0)\). Then \(BM=3\) and \(VM=(8-4,\,3-3,\,0-10)=(4,0,-10)\).
Since \(VM\perp BC\), \(\triangle BVM\) is right-angled at \(M\) and \(\angle BVC=2\angle BVM\).
In \(\triangle BVM\), \(VB=\sqrt{125}\) and \(BM=3\). Hence \[ \angle BVM=\arcsin\!\left(\frac{BM}{VB}\right) =\arcsin\!\left(\frac{3}{\sqrt{125}}\right) \approx 0.271657\ldots \] Therefore \(\displaystyle \angle BVC=2\angle BVM\approx 2(0.271657\ldots)=\boxed{0.543\ \text{rad}}\ (\approx \boxed{31.1^\circ}).\)
Equivalently, \(\displaystyle \angle BVM=\arccos\!\left(\frac{|VM|}{|VB|}\right)=\arccos\!\left(\frac{\sqrt{116}}{\sqrt{125}}\right)\), then double.
Total Marks: 6