Mathematics · Vector Algebra

JEE Main 2024 — 1 February, Shift 2 — Question 9

Consider a △ABC\triangle \mathrm{ABC} where A(1,3,2),B(−2,8,0)\mathrm{A}(1,3,2), \mathrm{B}(-2,8,0) and C(3,6,7)\mathrm{C}(3,6,7). If the angle bisector of ∠BAC\angle \mathrm{BAC} meets the line BC at D , then the length of the projection of the vector AD→\overrightarrow{A D} on the vector AC→\overrightarrow{A C} is:

  1. Option A:

    37238\frac{37}{2 \sqrt{38}}

    Correct
  2. Option B:

    382\frac{\sqrt{38}}{2}

  3. Option C:

    39238\frac{39}{2 \sqrt{38}}

  4. Option D:

    19\sqrt{19}

Answer: A

Step-by-step solution

A(1,3,2);B(−2,8,0);C(3,6,7)\mathrm{A}(1,3,2) ; \mathrm{B}(-2,8,0) ; \mathrm{C}(3,6,7)

AC→=2i^+3j^+5k^\overrightarrow{\mathrm{AC}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}

AB=9+25+4=38\mathrm{AB}=\sqrt{9+25+4}=\sqrt{38}

AC=4+9+25=38\mathrm{AC}=\sqrt{4+9+25}=\sqrt{38}

AD→=−12i^+4j^+32k^=−12(i^+8j^+3k^)\overrightarrow{\mathrm{AD}}=-\frac{1}{2} \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\frac{3}{2} \hat{\mathrm{k}}=-\frac{1}{2}(\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})

Length of projection of

AD→\overrightarrow{\mathrm{AD}} on AC→\overrightarrow{\mathrm{AC}}

=∣AD→⋅AC→∣AC→∣∣=37238=\left|\frac{\overrightarrow{\mathrm{AD}} \cdot \overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AC}}|}\right|=\frac{37}{2 \sqrt{38}}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Vector Algebra
Topic
Projection & component of a vector along another vector.
Consider a triangle ABC where A (1,3,2), B (-2,8,0) and C (3,6,7) .… | JEE Main 2024 PYQ with Solution · DhiX AI