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JEE Main 2018
Vector Algebra
Vector Algebra
Easy

Question

The vectors AB=3i^+4k^&AC=5i^2j^+4k^\overrightarrow {AB} = 3\widehat i + 4\widehat k\,\,\& \,\,\overrightarrow {AC} = 5\widehat i - 2\widehat j + 4\widehat k are the sides of triangle ABC.ABC. The length of the median through AA is :

Options

Solution

Key Concepts and Formulas

  • Vector Representation of a Median: If DD is the midpoint of side BCBC in triangle ABCABC, the vector representing the median from vertex AA to DD is given by AD=AB+AC2\overrightarrow{AD} = \frac{\overrightarrow{AB} + \overrightarrow{AC}}{2}. This formula arises from the midpoint formula for vectors and the triangle law of vector addition.
  • Magnitude of a Vector: The length or magnitude of a vector v=xi^+yj^+zk^\vec{v} = x\widehat{i} + y\widehat{j} + z\widehat{k} is calculated as v=x2+y2+z2|\vec{v}| = \sqrt{x^2 + y^2 + z^2}.

Step-by-Step Solution

Step 1: Identify the given side vectors. We are given the vectors representing two sides of triangle ABCABC originating from vertex AA: AB=3i^+4k^\overrightarrow{AB} = 3\widehat{i} + 4\widehat{k} AC=5i^2j^+4k^\overrightarrow{AC} = 5\widehat{i} - 2\widehat{j} + 4\widehat{k} To facilitate calculations, we can explicitly write the j^\widehat{j} component of AB\overrightarrow{AB} as 0: AB=3i^+0j^+4k^\overrightarrow{AB} = 3\widehat{i} + 0\widehat{j} + 4\widehat{k}

Step 2: Calculate the vector representing the median through A. The median through vertex AA connects AA to the midpoint DD of the side BCBC. We use the formula for the median vector: AD=AB+AC2\overrightarrow{AD} = \frac{\overrightarrow{AB} + \overrightarrow{AC}}{2} Substitute the given vectors into the formula: AD=(3i^+0j^+4k^)+(5i^2j^+4k^)2\overrightarrow{AD} = \frac{(3\widehat{i} + 0\widehat{j} + 4\widehat{k}) + (5\widehat{i} - 2\widehat{j} + 4\widehat{k})}{2} Add the corresponding components of the vectors: AD=(3+5)i^+(02)j^+(4+4)k^2\overrightarrow{AD} = \frac{(3+5)\widehat{i} + (0-2)\widehat{j} + (4+4)\widehat{k}}{2} AD=8i^2j^+8k^2\overrightarrow{AD} = \frac{8\widehat{i} - 2\widehat{j} + 8\widehat{k}}{2} Divide each component by 2 to obtain the median vector: AD=4i^j^+4k^\overrightarrow{AD} = 4\widehat{i} - \widehat{j} + 4\widehat{k}

Step 3: Calculate the length of the median. The length of the median is the magnitude of the vector AD\overrightarrow{AD}. Using the formula for the magnitude of a vector: AD=(4)2+(1)2+(4)2|\overrightarrow{AD}| = \sqrt{(4)^2 + (-1)^2 + (4)^2} AD=16+1+16|\overrightarrow{AD}| = \sqrt{16 + 1 + 16} AD=33|\overrightarrow{AD}| = \sqrt{33}

Common Mistakes & Tips

  • Missing Components: Be careful to include all components (i^\widehat{i}, j^\widehat{j}, k^\widehat{k}) in your vector addition, even if a component is zero. Forgetting the j^\widehat{j} component in AB\overrightarrow{AB} can lead to an incorrect sum.
  • Sign Errors: Pay close attention to the signs of the components when adding or subtracting vectors, especially when dealing with negative coefficients like 2j^-2\widehat{j}.
  • Squaring Errors: When calculating the magnitude, ensure that each component is squared correctly, and that negative signs are handled properly (e.g., (1)2=1(-1)^2 = 1).

Summary

To find the length of the median through vertex AA of a triangle ABCABC, we first determine the median vector AD\overrightarrow{AD} using the formula AD=AB+AC2\overrightarrow{AD} = \frac{\overrightarrow{AB} + \overrightarrow{AC}}{2}. Once the median vector is found by adding the components of AB\overrightarrow{AB} and AC\overrightarrow{AC} and dividing by two, its length is calculated by finding its magnitude using the formula AD=x2+y2+z2|\overrightarrow{AD}| = \sqrt{x^2 + y^2 + z^2}. In this case, the median vector is 4i^j^+4k^4\widehat{i} - \widehat{j} + 4\widehat{k}, and its length is 33\sqrt{33}.

The final answer is 33\boxed{\sqrt{33}} which corresponds to option (D).

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