Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Vector Algebra question

2021 · 18 Mar · Shift 2 · Q32
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Vector Algebra
  5. /2021 · 18 Mar · Shift 2 · Q32

Vector Algebra question

2021 · 18 Mar · Shift 2 · Q32

JEE MainMathematicsVector AlgebraMCQ+4 / −1
In a triangle ABC, if ∣BC→∣=8,∣CA→∣=7,∣AB→∣=10|\overrightarrow {BC} | = 8,|\overrightarrow {CA} | = 7,|\overrightarrow {AB} | = 10∣BC∣=8,∣CA∣=7,∣AB∣=10, then the projection of the vector AB→\overrightarrow {AB}AB on AC→\overrightarrow {AC}AC is equal to :
  1. A
    254{{25} \over 4}425​
  2. B
    12720{{127} \over 20}20127​
  3. C
    8514{{85} \over 14}1485​
  4. D
    11516{{115} \over 16}16115​
View written solutionFree

Correct answer: C

  1. Given side lengths of triangle ABCABCABC

We have: ∣BC→∣=8,∣CA→∣=7,∣AB→∣=10|\overrightarrow{BC}| = 8, \quad |\overrightarrow{CA}| = 7, \quad |\overrightarrow{AB}| = 10∣BC∣=8,∣CA∣=7,∣AB∣=10

We need the projection of AB→\overrightarrow{AB}AB on AC→\overrightarrow{AC}AC.

  1. Use dot product formula

The scalar projection of AB→\overrightarrow{AB}AB on AC→\overrightarrow{AC}AC is AB→⋅AC→∣AC→∣\frac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AC}|}∣AC∣AB⋅AC​

So first we find AB→⋅AC→\overrightarrow{AB}\cdot \overrightarrow{AC}AB⋅AC.

  1. Relate vectors in triangle

Since AB→+BC→=AC→\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}AB+BC=AC we get ∣BC→∣2=∣AC→−AB→∣2|\overrightarrow{BC}|^2 = |\overrightarrow{AC} - \overrightarrow{AB}|^2∣BC∣2=∣AC−AB∣2

Expanding, ∣BC→∣2=∣AC→∣2+∣AB→∣2−2 AB→⋅AC→|\overrightarrow{BC}|^2 = |\overrightarrow{AC}|^2 + |\overrightarrow{AB}|^2 - 2\,\overrightarrow{AB}\cdot\overrightarrow{AC}∣BC∣2=∣AC∣2+∣AB∣2−2AB⋅AC

Substitute the given magnitudes: 82=72+102−2 AB→⋅AC→8^2 = 7^2 + 10^2 - 2\,\overrightarrow{AB}\cdot\overrightarrow{AC}82=72+102−2AB⋅AC 64=49+100−2 AB→⋅AC→64 = 49 + 100 - 2\,\overrightarrow{AB}\cdot\overrightarrow{AC}64=49+100−2AB⋅AC 64=149−2 AB→⋅AC→64 = 149 - 2\,\overrightarrow{AB}\cdot\overrightarrow{AC}64=149−2AB⋅AC 2 AB→⋅AC→=852\,\overrightarrow{AB}\cdot\overrightarrow{AC} = 852AB⋅AC=85 AB→⋅AC→=852\overrightarrow{AB}\cdot\overrightarrow{AC} = \frac{85}{2}AB⋅AC=285​

  1. Find the projection

Since ∣AC→∣=7|\overrightarrow{AC}| = 7∣AC∣=7 therefore projection of AB→\overrightarrow{AB}AB on AC→\overrightarrow{AC}AC is

= \frac{\frac{85}{2}}{7} = \frac{85}{14}$$ 5. **Match with options** $$\frac{85}{14}$$ corresponds to **Option C**.
PreviousNext

More from Vector Algebra

  • Let a, b, c be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle θ, with the vector a+b+c…2021 · Numerical
  • If the shortest distance between the lines r1​​=αi+2j​+2k+λ(i−2j​+2k), λ∈ R, α> 0 and r2​​=−4i−k+μ(3i−2j​−2k)…2021 · Numerical
  • In a triangle ABC, if ​BC​=3, ​CA​=5 and ​BA​=7, then the projection of the vector BA on BC…2021 · MCQ
  • For p > 0, a vector v2​=2i+(p+1)j​ is obtained by rotating the vector v1​=3​pi+j​ by an angle θ about origin in counter clockwise…2021 · Numerical
  • Let a=i+2j​−k, b=i−j​ and c=i−j​−k be three given vectors. If r is a vector…2021 · Numerical
  • Let a=i+αj​+3k and b=3i−αj​+k. If the area of the parallelogram whose adjacent sides are represented by the vectors a…2021 · Numerical
  • Let p​=2i+3j​+k and q​=i+2j​+k be two vectors. If a vector r=(αi+βj​+γk)…2021 · Numerical
  • Let a, b and c be distinct positive numbers. If the vectors ai+aj​+ck,i+k and ci+cj​+bk are co-planar, then c is equal to :2021 · MCQ