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JEE Main 2023
Vector Algebra
Vector Algebra
Medium

Question

If the position vectors of the vertices A, B and C of a Δ\Delta ABC are respectively 4i^+7j^+8k^,4\widehat i + 7\widehat j + 8\widehat k, 2i^+3j^+4k^,2\widehat i + 3\widehat j + 4\widehat k, and 2i^+5j^+7k^,2\widehat i + 5\widehat j + 7\widehat k, then the position vectors of the point, where the bisector of \angle A meets BC is :

Options

Solution

  1. Key Concepts and Formulas
  • Angle Bisector Theorem: In a triangle ΔABC\Delta ABC, if the internal bisector of A\angle A intersects side BCBC at point DD, then the ratio of the lengths of the segments BDBD and DCDC is equal to the ratio of the lengths of the adjacent sides ABAB and ACAC. Mathematically, BDDC=ABAC\frac{BD}{DC} = \frac{AB}{AC}.
  • Section Formula (Internal Division): If a point DD divides the line segment joining points BB and CC internally in the ratio m:nm:n, then the position vector of DD, denoted by d\vec{d}, is given by d=nb+mcm+n\vec{d} = \frac{n\vec{b} + m\vec{c}}{m+n}, where b\vec{b} and c\vec{c} are the position vectors of BB and CC, respectively.
  1. Step-by-Step Solution
  • Step 1: Understand the Problem and Identify Given Information We are given the position vectors of the vertices AA, BB, and CC of a triangle ΔABC\Delta ABC: a=4i^+7j^+8k^\vec{a} = 4\widehat i + 7\widehat j + 8\widehat k b=2i^+3j^+4k^\vec{b} = 2\widehat i + 3\widehat j + 4\widehat k c=2i^+5j^+7k^\vec{c} = 2\widehat i + 5\widehat j + 7\widehat k We need to find the position vector of the point where the bisector of A\angle A meets side BCBC. Let this point be DD.

  • Step 2: Calculate the Vectors Representing Sides AB and AC To use the Angle Bisector Theorem, we first need the vectors representing the sides ABAB and ACAC. Vector AB=ba=(2i^+3j^+4k^)(4i^+7j^+8k^)\vec{AB} = \vec{b} - \vec{a} = (2\widehat i + 3\widehat j + 4\widehat k) - (4\widehat i + 7\widehat j + 8\widehat k) AB=(24)i^+(37)j^+(48)k^=2i^4j^4k^\vec{AB} = (2-4)\widehat i + (3-7)\widehat j + (4-8)\widehat k = -2\widehat i - 4\widehat j - 4\widehat k Vector AC=ca=(2i^+5j^+7k^)(4i^+7j^+8k^)\vec{AC} = \vec{c} - \vec{a} = (2\widehat i + 5\widehat j + 7\widehat k) - (4\widehat i + 7\widehat j + 8\widehat k) AC=(24)i^+(57)j^+(78)k^=2i^2j^1k^\vec{AC} = (2-4)\widehat i + (5-7)\widehat j + (7-8)\widehat k = -2\widehat i - 2\widehat j - 1\widehat k

  • Step 3: Calculate the Lengths of Sides AB and AC The lengths of the sides are the magnitudes of these vectors. Length of ABAB, denoted by ABAB or AB|\vec{AB}|: AB=AB=(2)2+(4)2+(4)2=4+16+16=36=6AB = |\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 Length of ACAC, denoted by ACAC or AC|\vec{AC}|: AC=AC=(2)2+(2)2+(1)2=4+4+1=9=3AC = |\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3

  • Step 4: Apply the Angle Bisector Theorem to Find the Ratio of Division According to the Angle Bisector Theorem, point DD divides the side BCBC in the ratio AB:ACAB:AC. BDDC=ABAC=63=21\frac{BD}{DC} = \frac{AB}{AC} = \frac{6}{3} = \frac{2}{1} Thus, point DD divides the line segment BCBC internally in the ratio m:n=2:1m:n = 2:1.

  • Step 5: Use the Section Formula to Find the Position Vector of D We use the section formula for internal division with m=2m=2, n=1n=1, b=2i^+3j^+4k^\vec{b} = 2\widehat i + 3\widehat j + 4\widehat k, and c=2i^+5j^+7k^\vec{c} = 2\widehat i + 5\widehat j + 7\widehat k. The position vector of DD, d\vec{d}, is given by: d=nb+mcm+n\vec{d} = \frac{n\vec{b} + m\vec{c}}{m+n} Substituting the values: d=1(2i^+3j^+4k^)+2(2i^+5j^+7k^)2+1\vec{d} = \frac{1 \cdot (2\widehat i + 3\widehat j + 4\widehat k) + 2 \cdot (2\widehat i + 5\widehat j + 7\widehat k)}{2+1}

  • Step 6: Perform Vector Addition and Scalar Multiplication First, perform the scalar multiplication: 1(2i^+3j^+4k^)=2i^+3j^+4k^1 \cdot (2\widehat i + 3\widehat j + 4\widehat k) = 2\widehat i + 3\widehat j + 4\widehat k 2(2i^+5j^+7k^)=4i^+10j^+14k^2 \cdot (2\widehat i + 5\widehat j + 7\widehat k) = 4\widehat i + 10\widehat j + 14\widehat k Now, add these resulting vectors: (2i^+3j^+4k^)+(4i^+10j^+14k^)=(2+4)i^+(3+10)j^+(4+14)k^(2\widehat i + 3\widehat j + 4\widehat k) + (4\widehat i + 10\widehat j + 14\widehat k) = (2+4)\widehat i + (3+10)\widehat j + (4+14)\widehat k =6i^+13j^+18k^= 6\widehat i + 13\widehat j + 18\widehat k Finally, divide by the sum of the ratio (m+n=3m+n=3): d=6i^+13j^+18k^3\vec{d} = \frac{6\widehat i + 13\widehat j + 18\widehat k}{3} d=13(6i^+13j^+18k^)\vec{d} = \frac{1}{3}(6\widehat i + 13\widehat j + 18\widehat k)

  1. Common Mistakes & Tips
  • Incorrect Ratio Assignment: Ensure that the ratio m:nm:n is correctly applied in the section formula. The term mm should be multiplied by the position vector of the point that is further from the dividing point (i.e., c\vec{c}), and nn by the position vector of the point closer to the dividing point (i.e., b\vec{b}), based on the ratio BD:DC=m:nBD:DC = m:n.
  • Sign Errors in Vector Subtraction: When calculating vectors like AB=ba\vec{AB} = \vec{b} - \vec{a}, pay close attention to the signs of the components.
  • Magnitude Calculation Errors: Double-check the squaring and summing of components when calculating the magnitude of vectors.
  1. Summary

The problem requires finding the position vector of the point where the angle bisector of A\angle A meets the side BCBC. This is achieved by first applying the Angle Bisector Theorem to determine the ratio in which the bisector divides BCBC, using the lengths of sides ABAB and ACAC. Subsequently, the Section Formula for internal division is used with the calculated ratio and the position vectors of BB and CC to find the required position vector. The calculated position vector is 13(6i^+13j^+18k^)\frac{1}{3}(6\widehat i + 13\widehat j + 18\widehat k).

The final answer is \boxed{\frac{1}{3}\left( {6\widehat i + 13\widehat j + 18\widehat k} \right)}. This corresponds to option (C).

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