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Chapter 2: Problem 165

Two points \(P\) and \(Q\) are given in the rectangular cartesian coordinates in the curve \(y=2^{x+2}\), such that OP. \(\hat{\mathbf{i}}=-1\) and \(\mathbf{O Q} \cdot \hat{\mathbf{i}}=2\), where \(\hat{\mathbf{i}}\) is a unit vector along the \(X\)-axis. Find the magnitude of \(\mathrm{OQ}-4 \mathrm{OP}\).

Short Answer

Expert verified
The magnitude of \( \mathrm{OQ} - 4\mathrm{OP} \) is 10.

Step by step solution

01

Determine the coordinates of Point P

The condition for point P is given as \( \mathbf{OP} \cdot \hat{\mathbf{i}}=-1 \). This means the x-coordinate of P is \(-1\). Substitute \(x = -1\) into the curve equation \(y = 2^{x+2}\):\[ y = 2^{(-1)+2} = 2^1 = 2 \]So the coordinates of P are \((-1, 2)\).
02

Determine the coordinates of Point Q

The condition for point Q is given as \( \mathbf{OQ} \cdot \hat{\mathbf{i}} = 2 \). This means the x-coordinate of Q is \(2\). Substitute \(x = 2\) into the curve equation \(y = 2^{x+2}\):\[ y = 2^{2+2} = 2^4 = 16 \]So the coordinates of Q are \((2, 16)\).
03

Calculate the vector OQ

The coordinates for Q are \((2, 16)\). Thus, the vector \( \mathbf{OQ} \) is \( (2, 16) \).
04

Calculate the vector OP

The coordinates for P are \((-1, 2)\). Thus, the vector \( \mathbf{OP} \) is \((-1, 2)\).
05

Calculate the vector OQ - 4OP

First, calculate \( 4 \mathbf{OP} \):\[ 4 imes (-1, 2) = (-4, 8) \]Now, compute \( \mathbf{OQ} - 4 \mathbf{OP} \):\[ (2, 16) - (-4, 8) = (2 + 4, 16 - 8) = (6, 8) \]
06

Find the magnitude of vector OQ - 4OP

The magnitude of a vector \( (x, y) \) is given by \( \sqrt{x^2 + y^2} \). Use this formula to find the magnitude of \((6, 8)\):\[ \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cartesian coordinates
In geometry, Cartesian coordinates are a system that uses two numbers to specify the location of a point on a plane. These numbers are known as the x-coordinate and the y-coordinate. The x-coordinate represents the horizontal position of the point on the x-axis, and the y-coordinate represents its vertical position on the y-axis. This system of coordinates is useful because it allows us to express geometric shapes and curves algebraically.
  • X-axis: The horizontal line along which the x-coordinate is measured.
  • Y-axis: The vertical line along which the y-coordinate is measured.
  • Origin: The point \(O(0,0)\) where both axes intersect.

In our problem, points P and Q are described using Cartesian coordinates. For example, point P has coordinates \((-1, 2)\), indicating it is 1 unit left of the origin on the x-axis and 2 units up on the y-axis.
Curve equation
A curve equation is a mathematical expression representing a set of points that form a curve on the coordinate plane. In general terms, a curve equation can illustrate profiles such as circles, ellipses, parabolas, or other complex shapes, depending on its formulation.

For this exercise, we use the curve equation \(y = 2^{x+2}\). This specific equation is an exponential function, indicating that as x changes, the value of y increases exponentially.

Let's see how it applies specifically:
  • When \(x = -1\), substituting into the curve yields \(y = 2^{1} = 2\), thus, point P is \((-1, 2)\).
  • When \(x = 2\), substituting yields \(y = 2^4 = 16\), yielding point Q as \( (2, 16) \).

This equation helps us identify the precise location of the points P and Q along the specified curve.
Vector magnitude
Vector magnitude represents the length or size of a vector. For a vector in the Cartesian plane, it is computed using the Pythagorean theorem. Given a vector with coordinates x and y, the magnitude is given by \( \sqrt{x^2 + y^2} \). This represents the distance of the vector point from the origin in the plane.

For this exercise, after calculating the vector \( \mathbf{OQ}-4\mathbf{OP} \), which is \( (6, 8) \), we find its magnitude using the formula:
\[ \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]
This calculation shows the distance and how much the vector \( \mathbf{OQ}-4\mathbf{OP} \) stretches in a geometric space.
Unit vector
A unit vector is a vector that has a magnitude of 1. It is often used to indicate direction without considering magnitude. In the coordinate plane, a standard example is the unit vector along each axis. For instance:
  • \( \hat{\mathbf{i}} \): The unit vector along the x-axis, with value (1, 0).
  • \( \hat{\mathbf{j}} \): The unit vector along the y-axis, with value (0, 1).
In the problem at hand, \( \hat{\mathbf{i}} \) signifies direction along the x-axis.

Vector operations such as dot product use unit vectors to understand vector alignments and projections. For example, the condition \( \mathbf{OP} \cdot \hat{\mathbf{i}} = -1 \) used here allows us to determine the x-component of point \(P\).

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Most popular questions from this chapter

If \(\alpha\) and \(\beta\) are two perpendicular unit vectors such that \(\mathbf{x}=\hat{\boldsymbol{\beta}}-(\alpha \times \mathbf{x})\), then the value of \(4|\mathbf{x}|^{2}\) is.

The vector \(\mathbf{a}=\alpha \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\beta \hat{\mathbf{k}}\) lies in the plane of the vectors \(\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and \(\mathbf{c}=\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and bisects the angle between \(\mathbf{b}\) and \(c\) Then, which one of the following gives \mathrm{\\{} p o s s i b l e ~ v a l u e s ~ o f ~ \(\alpha\) and \(\beta\) ? (a) \(\alpha=1, \beta=1\) (b) \(\alpha=2, \beta=2\) (c) \(\alpha=1, \beta=2\) (d) \(\alpha=2, \beta=1\)

Let \(\mathbf{a}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\) and \(\mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) be three vectors. A vector in the plane of \(\mathbf{b}\) and \(c\) whose projection on \(\mathrm{a}\) is of magnitude \(\sqrt{\frac{2}{3}}\) is (a) \(2 \hat{i}+3 \hat{j}-3 \hat{\mathbf{k}}\) (b) \(2 \hat{i}+3 \hat{j}+3 \hat{k}\) (c) \(-2 \hat{i}-\hat{j}+5 \hat{\mathbf{k}}\) (d) \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}}\)

The distance between the line \(\mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}})\) and the plane \(\mathbf{r} \cdot(\hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\overrightarrow{\mathbf{k}})=5\) is \(\quad\) [AIEEE 2005] (a) \(\frac{10}{3}\) (b) \(\frac{3}{10}\) (c) \(\frac{10}{3 \sqrt{3}}\) (d) \(\frac{10}{9}\)

If the vectors \(\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\mathbf{c}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mu \hat{\mathbf{k}}\) are mutually orthogonal, then \((\lambda, \mu)\) is equal to [AJEEE 2010] (a) \((-3,2)\) (b) \((2,-3)\) (c) \((-2,3)\) (d) \((3,-2)\)
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