Problem 2
Question
Let the position vectors of points ' \(A\) ' and ' \(B^{\prime}\) be \(\hat{i}+\hat{j}+\hat{k}\) and \(2 \hat{i}+\hat{j}+3 \hat{k}\), respectively. A point ' \(\mathrm{P}\) ' divides the line segment \(\mathrm{AB}\) internally in the ratio \(\lambda: 1(\lambda>0)\). If \(O\) is the region and \(\overrightarrow{O B} \cdot \overrightarrow{O P}-3|\overrightarrow{O A} \times \overrightarrow{O P}|^{2}=6\), then \(\lambda\) is equal to
Step-by-Step Solution
Verified Answer
\( \lambda = \frac{-8 + 4\sqrt{22}}{16}\).
1Step 1: Understand the Problem and Given Vectors
We need to find the scalar \( \lambda \) that divides the line segment \( AB \) internally in the ratio \( \lambda : 1 \). We are given the position vectors \( \overrightarrow{OA} = \hat{i} + \hat{j} + \hat{k} \) and \( \overrightarrow{OB} = 2\hat{i} + \hat{j} + 3\hat{k} \). Additionally, the condition \( \overrightarrow{OB} \cdot \overrightarrow{OP} - 3| \overrightarrow{OA} \times \overrightarrow{OP} |^2 = 6 \) must hold.
2Step 2: Find the Position Vector of Point P
The point \( P \) divides \( AB \) in the ratio \( \lambda : 1 \), so its position vector is given by:\[\overrightarrow{OP} = \frac{\lambda \cdot \overrightarrow{OB} + 1 \cdot \overrightarrow{OA}}{\lambda + 1}\].Plug in the given vectors:\[\overrightarrow{OP} = \frac{\lambda(2\hat{i} + \hat{j} + 3\hat{k}) + (\hat{i} + \hat{j} + \hat{k})}{\lambda + 1}\].Simplify:\[\overrightarrow{OP} = \frac{(2\lambda + 1)\hat{i} + (\lambda + 1)\hat{j} + (3\lambda + 1)\hat{k}}{\lambda + 1}\].
3Step 3: Calculate \( \overrightarrow{OB} \cdot \overrightarrow{OP} \)
Find the dot product \( \overrightarrow{OB} \cdot \overrightarrow{OP} \):ewline\(\overrightarrow{OB} = 2\hat{i} + \hat{j} + 3\hat{k}\). \(\overrightarrow{OP} = \frac{(2\lambda + 1)\hat{i} + (\lambda + 1)\hat{j} + (3\lambda + 1)\hat{k}}{\lambda + 1}\). Calculate the dot product:\[\overrightarrow{OB} \cdot \overrightarrow{OP} = 2 \left(\frac{2\lambda + 1}{\lambda + 1}\right) + \frac{\lambda + 1}{\lambda + 1} + 3 \left(\frac{3\lambda + 1}{\lambda + 1}\right)\].Simplify:\[= \frac{4\lambda + 2 + \lambda + 1 + 9\lambda + 3}{\lambda + 1} = \frac{14\lambda + 6}{\lambda + 1}\].
4Step 4: Calculate \( | \overrightarrow{OA} \times \overrightarrow{OP} |^2 \)
First find \( \overrightarrow{OA} \times \overrightarrow{OP} \):\( \overrightarrow{OA} = \hat{i} + \hat{j} + \hat{k} \).Using the cross product formula:\(\overrightarrow{OA} \times \overrightarrow{OP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ \frac{2\lambda + 1}{\lambda + 1} & \frac{\lambda + 1}{\lambda + 1} & \frac{3\lambda + 1}{\lambda + 1} \end{vmatrix}\).Calculate the determinant:\(= \hat{i}(1 \cdot \frac{3\lambda + 1}{\lambda + 1} - 1) - \hat{j}(\frac{2\lambda + 1}{\lambda + 1} - \frac{3\lambda + 1}{\lambda + 1}) + \hat{k}(\frac{2\lambda + 1}{\lambda + 1} - 1)\).Simplify: \(= \frac{2\lambda}{\lambda + 1}\hat{i} - \frac{\lambda}{\lambda + 1}\hat{j} + \frac{\lambda + 1}{\lambda + 1}\hat{k}\).Then calculate magnitude squared:\(| \overrightarrow{OA} \times \overrightarrow{OP} |^2 = \left(\frac{2\lambda}{\lambda + 1}\right)^2 + \left(-\frac{\lambda}{\lambda + 1}\right)^2 + \left(\frac{1}{\lambda + 1}\right)^2\).Simplify:\[= \frac{4\lambda^2 + \lambda^2 + 1}{(\lambda + 1)^2} = \frac{5\lambda^2 + 1}{(\lambda + 1)^2}\].
5Step 5: Formulate and Solve the Equation
Using the given condition \(\overrightarrow{OB} \cdot \overrightarrow{OP} - 3|\overrightarrow{OA} \times \overrightarrow{OP}|^2 = 6\), substitute the expressions calculated:\[\frac{14\lambda + 6}{\lambda + 1} - 3 \cdot \frac{5\lambda^2 + 1}{(\lambda + 1)^2} = 6\].Clear the fractions by multiplying through by \((\lambda + 1)^2\):\[14\lambda(\lambda + 1) + 6(\lambda + 1) - 3(5\lambda^2 + 1) = 6(\lambda + 1)^2\].Expand and simplify:\[14\lambda^2 + 14\lambda + 6\lambda + 6 - 15\lambda^2 - 3 = 6\lambda^2 + 12\lambda + 6\].Simplify and combine like terms:\[14\lambda^2 + 20\lambda + 3 = 6\lambda^2 + 12\lambda + 12\].Solve the quadratic equation in standard form and find the positive \(\lambda\).
6Step 6: Solve the Quadratic Equation
Rearrange and simplify the equation to standard form:\[8\lambda^2 + 8\lambda - 9 = 0\].Use the quadratic formula: \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 8\), \(b = 8\), and \(c = -9\).Calculate the discriminant:\[b^2 - 4ac = 8^2 - 4(8)(-9) = 64 + 288 = 352\].Substitute and solve:\[\lambda = \frac{-8 \pm \sqrt{352}}{16}\], \[\sqrt{352} = 2\sqrt{88} = 2 \times 2 \times \sqrt{22} = 4\sqrt{22}\]Thus the solutions for \(\lambda\):\[\lambda = \frac{-8 + 4\sqrt{22}}{16}, \frac{-8 - 4\sqrt{22}}{16}\].Select the positive value:\[\lambda = \frac{-8 + 4\sqrt{22}}{16}\].
Key Concepts
Position VectorsLine Segment DivisionCross ProductDot Product
Position Vectors
Position vectors are fundamental in understanding points in space. They describe the location of a point with respect to an origin point, usually denoted as \( O \). For instance, in our problem, the position vectors for points \( A \) and \( B \) are \( \overrightarrow{OA} = \hat{i} + \hat{j} + \hat{k} \) and \( \overrightarrow{OB} = 2\hat{i} + \hat{j} + 3\hat{k} \) respectively. This notation uses the unit vectors \( \hat{i}, \hat{j}, \) and \( \hat{k} \) to represent directions along the x, y, and z axes, respectively.
Position vectors are critical because they allow for the representation of points in a three-dimensional plane using Cartesian coordinates. This becomes especially useful when calculating distances or specific conditions between points, like in the given exercise. To find a point \( P \) that divides a line, understanding and manipulating these vectors is necessary.
Naturally, each position vector gives very precise information, contributing to the accurate resolution of problems involving spatial relationships.
Position vectors are critical because they allow for the representation of points in a three-dimensional plane using Cartesian coordinates. This becomes especially useful when calculating distances or specific conditions between points, like in the given exercise. To find a point \( P \) that divides a line, understanding and manipulating these vectors is necessary.
Naturally, each position vector gives very precise information, contributing to the accurate resolution of problems involving spatial relationships.
Line Segment Division
The concept of a point dividing a line segment is essential in vector algebra. In our exercise, the point \( P \) divides the line segment between \( A \) and \( B \) in an internal ratio \( \lambda : 1 \). This means \( P \) is closer to \( A \) by a factor determined by \( \lambda \). The formula for this division is crucial: \[ \overrightarrow{OP} = \frac{\lambda \cdot \overrightarrow{OB} + 1 \cdot \overrightarrow{OA}}{\lambda + 1} \] This helps to find the exact position vector for \( P \).
Understanding this is key because line segment division often forms the basis for more complex vector calculations. By using ratios, you can determine how far or close \( P \) is from either endpoint. It’s a fundamental concept that bridges geometry and algebra.
Understanding this is key because line segment division often forms the basis for more complex vector calculations. By using ratios, you can determine how far or close \( P \) is from either endpoint. It’s a fundamental concept that bridges geometry and algebra.
Cross Product
The cross product of two vectors is a vector perpendicular to both, usually in three-dimensional space. In this problem, the cross product \( \overrightarrow{OA} \times \overrightarrow{OP} \) is calculated to understand the spatial relationship between line segments. The cross product is particularly significant because its magnitude gives the area of the parallelogram formed by two vectors.
Here, we compute the cross product using the determinant method:
\[ \overrightarrow{OA} \times \overrightarrow{OP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ \frac{2\lambda + 1}{\lambda + 1} & \frac{\lambda + 1}{\lambda + 1} & \frac{3\lambda + 1}{\lambda + 1} \end{vmatrix} \]
Yet, it’s the magnitude squared of this cross product that comes into play, exploring the relationship:
\[ |\overrightarrow{OA} \times \overrightarrow{OP}|^2 = \left(\frac{2\lambda}{\lambda + 1}\right)^2 + \left(-\frac{\lambda}{\lambda + 1}\right)^2 + \left(\frac{1}{\lambda + 1}\right)^2 \]
This expansion helps solve constraints laid out in the problem, highlighting the relevance of cross product in determining alignment and distances indirectly.
Here, we compute the cross product using the determinant method:
\[ \overrightarrow{OA} \times \overrightarrow{OP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ \frac{2\lambda + 1}{\lambda + 1} & \frac{\lambda + 1}{\lambda + 1} & \frac{3\lambda + 1}{\lambda + 1} \end{vmatrix} \]
Yet, it’s the magnitude squared of this cross product that comes into play, exploring the relationship:
\[ |\overrightarrow{OA} \times \overrightarrow{OP}|^2 = \left(\frac{2\lambda}{\lambda + 1}\right)^2 + \left(-\frac{\lambda}{\lambda + 1}\right)^2 + \left(\frac{1}{\lambda + 1}\right)^2 \]
This expansion helps solve constraints laid out in the problem, highlighting the relevance of cross product in determining alignment and distances indirectly.
Dot Product
The dot product is another essential vector operation, providing a quick scalar result that tells us about the angle between two vectors and their magnitudes. It is used in this problem to relate the vectors of points \( O \), \( B \), and \( P \). The formula is:
\[ \overrightarrow{OB} \cdot \overrightarrow{OP} = (2\hat{i} + \hat{j} + 3\hat{k}) \cdot \left( \frac{(2\lambda + 1)\hat{i} + (\lambda + 1)\hat{j} + (3\lambda + 1)\hat{k}}{\lambda + 1} \right) \]
The solution for this scalar provides insight into the projection of one vector onto another, showing how aligned they are relative to the origin. This calculation helps in satisfying conditions specified in vector equations, like in this problem. The dot product is crucial for understanding relationships between vectors in multi-dimensional spaces, making it a valuable tool in the resolution of equations involving geometry and physics.
\[ \overrightarrow{OB} \cdot \overrightarrow{OP} = (2\hat{i} + \hat{j} + 3\hat{k}) \cdot \left( \frac{(2\lambda + 1)\hat{i} + (\lambda + 1)\hat{j} + (3\lambda + 1)\hat{k}}{\lambda + 1} \right) \]
The solution for this scalar provides insight into the projection of one vector onto another, showing how aligned they are relative to the origin. This calculation helps in satisfying conditions specified in vector equations, like in this problem. The dot product is crucial for understanding relationships between vectors in multi-dimensional spaces, making it a valuable tool in the resolution of equations involving geometry and physics.
Other exercises in this chapter
Problem 1
Let \(a, b, c \in \mathbf{R}\) be such that \(a^{2}+b^{2}+c^{2}=1\). If \(a \cos \theta-b \cos \left(\theta+\frac{2 \pi}{3}\right)=c \cos \left(\theta+\frac{4 \
View solution Problem 3
If the vectors, \(\bar{p}=(a+1) \hat{i}+a \hat{j}+a \hat{k}, \vec{q}=a \hat{i}+(a+1) \hat{j}+a \hat{k}\) and \(\vec{r}=a \hat{i}+a \hat{j}+(a+1) \hat{k} \quad(a
View solution Problem 4
Let \(\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\) be two vectors. If a vector perpendicular to both the vectors \(\vec{a
View solution Problem 5
If the volume of parallelopiped formed by the vectors \(\hat{i}+\lambda \hat{j}+\hat{k}, \hat{j}+\lambda \hat{k}\) and \(\lambda \hat{i}+\hat{k}\) is minimum, t
View solution