Vectors & 3D GeometrymediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Vectors & 3D Geometry: Let Origin Equals (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let OO be the origin with OP→=a⃗\overrightarrow{OP}=\vec{a} and OQ→=b⃗\overrightarrow{OQ}=\vec{b}. If RR is on OP→\overrightarrow{OP} such that OP→=5OR→\overrightarrow{OP}=5\overrightarrow{OR}, and MM is such that OQ→=5RM→\overrightarrow{OQ}=5\overrightarrow{RM}, then PM→\overrightarrow{PM} equals:
A15(a⃗−4b⃗)\dfrac{1}{5}(\vec{a}-4\vec{b})
B15(b⃗−4a⃗)\dfrac{1}{5}(\vec{b}-4\vec{a})correct
C15(−a⃗+4b⃗)\dfrac{1}{5}(-\vec{a}+4\vec{b})
D15(−b⃗+4a⃗)\dfrac{1}{5}(-\vec{b}+4\vec{a})
Solution
Step 1: Express OR→\overrightarrow{OR}
OP→=5OR→  ⟹  OR→=a⃗5\overrightarrow{OP} = 5\overrightarrow{OR} \implies \overrightarrow{OR} = \frac{\vec{a}}{5}
 Step 2: Express OM→\overrightarrow{OM}
OQ→=5RM→  ⟹  RM→=b⃗5\overrightarrow{OQ} = 5\overrightarrow{RM} \implies \overrightarrow{RM} = \frac{\vec{b}}{5}
OM→=OR→+RM→=a⃗5+b⃗5=a⃗+b⃗5\overrightarrow{OM} = \overrightarrow{OR}+\overrightarrow{RM} = \frac{\vec{a}}{5}+\frac{\vec{b}}{5} = \frac{\vec{a}+\vec{b}}{5}
 Step 3: Compute PM→\overrightarrow{PM}
PM→=OM→−OP→=a⃗+b⃗5−a⃗=a⃗+b⃗−5a⃗5=15(b⃗−4a⃗)\overrightarrow{PM} = \overrightarrow{OM}-\overrightarrow{OP} = \frac{\vec{a}+\vec{b}}{5}-\vec{a} = \frac{\vec{a}+\vec{b}-5\vec{a}}{5} = \frac{1}{5}(\vec{b}-4\vec{a})
 Answer: (2)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Vectors & 3D Geometry · medium
If the distance of the point (a,2,5)(a,2,5) from the image of (1,2,7)(1,2,7) in the line x1=y−11=z−22\dfrac{x}{1}=\dfrac{y-1}{1}=\dfrac{z-2}{2}is 4, then the sum of all possible values of aa is:
Vectors & 3D Geometry · medium
A line with direction ratios 1,−1,21,-1,2 intersects the lines x2=y3=z+13\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z+1}{3} and x+1−1=y−21=z4\dfrac{x+1}{-1}=\dfrac{y-2}{1}=\dfrac{z}{4} at the points PP and QQ respectively. If the length of the line segment PQPQ is α\alpha, then 225α2225\alpha^2 is equal to
Vectors & 3D Geometry · medium
Let △PQR\triangle PQR be such that PP and QQ lie on x+38=y−42=z+12\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2} and are at distance 6 from R(1,2,3)R(1,2,3). If (α,β,γ)(\alpha,\beta,\gamma) is the centroid of △PQR\triangle PQR, then α+β+γ\alpha+\beta+\gamma equals:
Vectors & 3D Geometry · medium
Let the vectors a⃗=−i^+j^+3k^\vec a=-\hat i+\hat j+3\hat k and b⃗=i^+3j^+k^\vec b=\hat i+3\hat j+\hat k. For some λ,μ∈R\lambda,\mu\in\mathbb{R}, let c⃗=λa⃗+μb⃗\vec c=\lambda\vec a+\mu\vec b. If c⃗⋅(3i^−6j^+2k^)=10\vec c\cdot(3\hat i-6\hat j+2\hat k)=10 and c⃗⋅(i^+j^+k^)=−2\vec c\cdot(\hat i+\hat j+\hat k)=-2, then ∣c⃗∣2|\vec c|^2 is equal to

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath