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

2025 · 24 Jan · Shift 2 · Q35
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. /2025 · 24 Jan · Shift 2 · Q35

Vector Algebra question

2025 · 24 Jan · Shift 2 · Q35

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let the position vectors of three vertices of a triangle be 4p⃗+q⃗−3r⃗,−5p⃗+q⃗+2r⃗4 \vec{p}+\vec{q}-3 \vec{r},-5 \vec{p}+\vec{q}+2 \vec{r}4p​+q​−3r,−5p​+q​+2r and 2p⃗−q⃗+2r⃗2 \vec{p}-\vec{q}+2 \vec{r}2p​−q​+2r. If the position vectors of the orthocenter and the circumcenter of the triangle are p⃗+q⃗+r⃗4\frac{\vec{p}+\vec{q}+\vec{r}}{4}4p​+q​+r​ and αp⃗+βq⃗+γr⃗\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r}αp​+βq​+γr respectively, then α+2β+5γ\alpha+2 \beta+5 \gammaα+2β+5γ is equal to :
  1. A
    4
  2. B
    3
  3. C
    1
  4. D
    6
View written solutionFree

Correct answer: B

Let the three vertices of the triangle be A=4p⃗+q⃗−3r⃗,B=−5p⃗+q⃗+2r⃗,C=2p⃗−q⃗+2r⃗.A=4\vec p+\vec q-3\vec r,\quad B=-5\vec p+\vec q+2\vec r,\quad C=2\vec p-\vec q+2\vec r.A=4p​+q​−3r,B=−5p​+q​+2r,C=2p​−q​+2r.

We are given the orthocenter H=p⃗+q⃗+r⃗4H=\frac{\vec p+\vec q+\vec r}{4}H=4p​+q​+r​ and the circumcenter O=αp⃗+βq⃗+γr⃗.O=\alpha \vec p+\beta \vec q+\gamma \vec r.O=αp​+βq​+γr.

We use the vector identity for a triangle:

If a⃗,b⃗,c⃗\vec a,\vec b,\vec ca,b,c are the position vectors of the vertices, and h⃗,o⃗\vec h,\vec oh,o are those of the orthocenter and circumcenter, then h⃗=a⃗+b⃗+c⃗−2o⃗.\vec h=\vec a+\vec b+\vec c-2\vec o.h=a+b+c−2o. Equivalently, o⃗=a⃗+b⃗+c⃗−h⃗2.\vec o=\frac{\vec a+\vec b+\vec c-\vec h}{2}.o=2a+b+c−h​.

1. Compute A+B+CA+B+CA+B+C

Add the coefficients of p⃗,q⃗,r⃗\vec p,\vec q,\vec rp​,q​,r separately:

For p⃗\vec pp​: 4+(−5)+2=14+(-5)+2=14+(−5)+2=1

For q⃗\vec qq​: 1+1+(−1)=11+1+(-1)=11+1+(−1)=1

For r⃗\vec rr: −3+2+2=1-3+2+2=1−3+2+2=1

Hence, A+B+C=p⃗+q⃗+r⃗.A+B+C=\vec p+\vec q+\vec r.A+B+C=p​+q​+r.

2. Use the orthocenter-circumcenter relation

Since H=p⃗+q⃗+r⃗4,H=\frac{\vec p+\vec q+\vec r}{4},H=4p​+q​+r​, we get O=(p⃗+q⃗+r⃗)−p⃗+q⃗+r⃗42.O=\frac{(\vec p+\vec q+\vec r)-\frac{\vec p+\vec q+\vec r}{4}}{2}.O=2(p​+q​+r)−4p​+q​+r​​.

Simplify inside:

=\frac{3}{4}(\vec p+\vec q+\vec r).$$ So, $$O=\frac{1}{2}\cdot \frac{3}{4}(\vec p+\vec q+\vec r) =\frac{3}{8}(\vec p+\vec q+\vec r).$$ Therefore, $$\alpha=\beta=\gamma=\frac{3}{8}.$$ --- ### 3. Compute $\alpha+2\beta+5\gamma$ $$\alpha+2\beta+5\gamma =\frac{3}{8}+2\cdot\frac{3}{8}+5\cdot\frac{3}{8}$$ $$=\frac{3}{8}(1+2+5)=\frac{3}{8}\cdot 8=3.$$ --- ### 4. Final answer $$\boxed{3}$$ So the correct option is **B**. --- ### 5. Comparison with stored answer Stored correct answer: **B** Our derived answer is also **B**, so they agree.
PreviousNext

More from Vector Algebra

  • Let a=3i^−j^​+2k^, b=a×(i^−2k^) and c=b×k^. Then the…2025 · MCQ
  • Let a=i^+j^​+k^,b=2i^+2j^​+k^ and d=a×b. If c…2025 · Numerical
  • If the components of a=αi^+βj^​+γk^ along and perpendicular to b=3i^+j^​−k^ respectively, are 1116​(3i^+j^​−k^) and 111​(−4i^−5j^​−17k^)…2025 · MCQ
  • Let A,B,C be three points in xy-plane, whose position vector are given by 3​i^+j^​,i^+3​j^​ and ai^+(1−a)j^​ respectively with respect to the origin O . If the distance of the point C…2025 · MCQ
  • Let a=i^+2j^​+k^ and b=2i^+7j^​+3k^. Let L1​:r=(−i^+2j^​+k^)+λa,λ∈R and L2​:r=(j^​+k^)+μb,μ∈R…2025 · MCQ
  • Let a=2i^−j^​+3k^, b=3i^−5j^​+k^ and c be a vector such that a×c=a×b=c×b and (a+c)⋅(b+c)=168…2025 · MCQ
  • Let a^ be a unit vector perpendicular to the vectors b=i^−2j^​+3k^ and c=2i^+3j^​−k^, and a^ makes an angle of cos−1(−31​) with the vector i^+j^​+k^…2025 · MCQ
  • Let a=−5i^+j^​−3k^,b=i^+2j^​−4k^ and c=(((a×b)×i^)×i^)×i^…2024 · MCQ