Vector Algebra
Angle bisector and section formula
Grade 12
Question:
<p>If the position vectors of the vertices <em>A</em>, <em>B</em> and <em>C</em> of a \(\triangle ABC\) are, respectively, \(4\hat{i}+7\hat{j}+8\hat{k}\), \(2\hat{i}+3\hat{j}+4\hat{k}\) and \(2\hat{i}+5\hat{j}+7\hat{k}\), then the position vector of the point, where the bisector of \(\angle A\) meets <em>BC</em> is</p>
<p>\(\dfrac{1}{2}(4\hat{i}+8\hat{j}+11\hat{k})\)</p>
<p>\(\dfrac{1}{3}(6\hat{i}+11\hat{j}+15\hat{k})\)</p>
<p>\(\dfrac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})\)</p>
<p>\(\dfrac{1}{4}(8\hat{i}+14\hat{j}+19\hat{k})\)</p>
Step-by-Step Solution
Key Concept: The angle bisector theorem states that the angle bisector divides the opposite side in the ratio of the adjacent sides. Use |AB| : |AC| to find the ratio in which the bisector meets BC, then apply section formula.
Step 1: Find vectors AB and AC A = 4î + 7ĵ + 8k̂, B = 2î + 3ĵ + 4k̂, C = 2î + 5ĵ + 7k̂ AB = B - A = -2î - 4ĵ - 4k̂ AC = C - A = -2î - 2ĵ - k̂ Step 2: Calculate magnitudes |AB| and |AC| |AB| = √(4 + 16 + 16) = √36 = 6 |AC| = √(4 + 4 + 1) = √9 = 3 Step 3: Apply angle bisector theorem The angle bisector from A meets BC at point D, dividing it in ratio |AB| : |AC| = 6 : 3 = 2 : 1 So D divides BC internally in ratio 2:1 Step 4: Apply section formula Position vector of D = (2·C + 1·B)/(2 + 1) = (2(2î + 5ĵ + 7k̂) + (2î + 3ĵ + 4k̂))/3 = (4î + 10ĵ + 14k̂ + 2î + 3ĵ + 4k̂)/3 = (6î + 13ĵ + 18k̂)/3 = 2î + (13/3)ĵ + 6k̂ ∴ Answer: C
Correct Answer: C