Question

If \(P,Q\) are two points on the curve \(y=2^{x+2}\) in the rectangular Cartesian coordinate system such that \(\overrightarrow{OP}\cdot \vec{i}=-1,\;\overrightarrow{OQ}\cdot \vec{i}=2\), then \(\overrightarrow{OQ}-4\overrightarrow{OP}=\)

• A. \(3\vec{i}+8\vec{j}\)
• B. \(4\vec{i}+6\vec{j}\)
• C. \(6\vec{i}+8\vec{j}\)
• D. \(4\vec{i}+3\vec{j}\)

Hint:

For a position vector \(\overrightarrow{OP}=x\vec{i}+y\vec{j}\), the value of \(\overrightarrow{OP}\cdot \vec{i}\) gives the \(x\)-coordinate of the point \(P\).

Answer 1

The Correct Option is C

Step 1: Understand the dot products.
The condition $\overrightarrow{OP}\cdot\vec i$ gives the $x$-coordinate of $P$, and similarly for $Q$. The points lie on $y=2^{x+2}$.
Step 2: Find point $P$.
Here $x=-1$, so $y=2^{-1+2}=2^1=2$. Thus $P=(-1,2)$ and $\overrightarrow{OP}=-\vec i+2\vec j$.
Step 3: Find point $Q$.
Here $x=2$, so $y=2^{2+2}=2^4=16$. Thus $Q=(2,16)$ and $\overrightarrow{OQ}=2\vec i+16\vec j$.
Step 4: Form the combination.
We need $\overrightarrow{OQ}-4\overrightarrow{OP}$.
Step 5: Compute component-wise.
\[ (2\vec i+16\vec j)-4(-\vec i+2\vec j)=(2+4)\vec i+(16-8)\vec j. \] Step 6: Simplify and conclude.
This gives $6\vec i+8\vec j$.
\[ \boxed{6\vec i+8\vec j} \]