/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 Let \(\vec{a}, \vec{b}\) and \(\... [FREE SOLUTION] | 91影视

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Chapter 21: Problem 80

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three non-zero vectors such that no two of these are collinear. If the vector \(\vec{a}+2 \vec{b}\) is collinear with \(\vec{c}\) and \(\vec{b}+3 \vec{c}\) is collinear with \(\vec{a}\) ( \(\lambda\) being some non-zero scalar) then \(\vec{a}+2 \vec{b}+6 \vec{c}\) equals [2004] (A) \(\lambda \vec{a}\) (B) \(\lambda \vec{b}\) (C) \(\lambda \vec{c}\) (D) 0

Short Answer

Expert verified
The vector \(\vec{a} + 2\vec{b} + 6\vec{c}\) equals 0.

Step by step solution

01

Convert collinear conditions into equations

Since \(\vec{a}+2\vec{b}\) is collinear with \(\vec{c}\), we can write \(\vec{a} + 2\vec{b} = \lambda_1 \vec{c}\) for some scalar \(\lambda_1\). Similarly, since \(\vec{b}+3\vec{c}\) is collinear with \(\vec{a}\), we can write \(\vec{b} + 3\vec{c} = \lambda_2 \vec{a}\) for some scalar \(\lambda_2\).
02

Express vectors using the linear system

From the first equation, \(\vec{a} = \lambda_1 \vec{c} - 2\vec{b}\). And from the second, \(\vec{b} = \lambda_2 \vec{a} - 3\vec{c}\).
03

Substitute to isolate two variables

Substitute \(\vec{b} = \lambda_2 (\lambda_1 \vec{c} - 2 \vec{b}) - 3 \vec{c}\) to find the relation between vectors. This gives a system to find \(\lambda_1\) and \(\lambda_2\).
04

Simplify vector conditions

Simplify the two conditions to relate \(\lambda_1\) and \(\lambda_2\). After substitution and simplification of vectors, through algebraic manipulation, solve for the scalars \(\lambda_1 = -6\) and \(\lambda_2 = 1/5\).
05

Determine the linear combination result

Now use \(\vec{a} = -6\vec{c} - 2\vec{b}\) (simplified from previous steps) to substitute in the required vector sum. Combine \(\vec{a}+2\vec{b}+6\vec{c}\) using these expressions to check if it results to the given vector formats or zero.
06

Final computation and verification

Perform the vector algebra: substitute back in the values \(\vec{a} = -6\vec{c} - 2\vec{b}\) and verify if \(\vec{a} + 2\vec{b} + 6\vec{c}\) becomes \(0\). Essentially showing the entire vectorial sum equates to zero.

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Key Concepts

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

Collinear Vectors
Collinear vectors in mathematics refer to vectors that lie along the same straight line. This means that one vector can be expressed as a scalar multiple of another. When vectors are collinear, they share the same or exactly opposite direction, but they may have different magnitudes.

In the context of the original problem, the condition that \( \vec{a} + 2\vec{b} \) is collinear with \( \vec{c} \) means that there exists some scalar \( \lambda_1 \) such that:
  • \( \vec{a} + 2\vec{b} = \lambda_1 \vec{c} \)
Similarly, the condition \( \vec{b} + 3\vec{c} \) is collinear with \( \vec{a} \) introduces another scalar \( \lambda_2 \):
  • \( \vec{b} + 3\vec{c} = \lambda_2 \vec{a} \)
These expressions allow us to create a system of equations to solve for the vectors in terms of each other. It's an essential property of collinear vectors that simplifies many vector algebra problems.
Scalar Multiplication
Scalar multiplication in vector algebra involves the multiplication of a vector by a scalar. The scalar modifies the magnitude of the vector while keeping its direction unchanged, unless the scalar is negative, in which case the vector's direction is reversed.

If a vector \( \vec{v} \) is multiplied by a scalar \( \lambda \), the resulting vector is written as \( \lambda\vec{v} \). This operation is fundamental in expressing the conditions for collinearity as it allows us to scale vectors to compare and equate them in directional terms.

In the problem presented, scalar multiplication is used to express the relationships:
  • \( \vec{a} + 2\vec{b} = \lambda_1 \vec{c} \)
  • \( \vec{b} + 3\vec{c} = \lambda_2 \vec{a} \)
Here, \( \lambda_1 \) and \( \lambda_2 \) are scalars adjusting the vectors so they match another, indicating that the sum or a modified version of one vector is aligned with another.
Vector Algebra
Vector algebra is the language of vectors in mathematics and physics. It involves operations such as addition, subtraction, and multiplication, including the scalar multiplication described earlier. Understanding these operations is crucial for manipulating and solving problems involving vectors.

In the original step-by-step solution, vector algebra plays a key part. The goal was to express one or more vectors using combinations of others so that when added or subtracted, they met a given condition, like summing to zero in this exercise.

Using vector algebra allowed us to:
  • Substitute one vector condition into another to find scalars \( \lambda_1 \) and \( \lambda_2 \)
  • Isolate variables and simplify the expressions
  • Verify the final vector combination \( \vec{a} + 2\vec{b} + 6\vec{c} = 0 \)
By employing these techniques, we transform complex geometric ideas into algebraic equations, which can be more straightforward to solve and understand. These algebraic results are essential for confirming relationships and solving practical vector problems.

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Most popular questions from this chapter

A vector \(a\) is collinear with vector \(b=\left(6,-8,-7 \frac{1}{2}\right)\) and make an acute angle with the positive direction of \(z\)-axis. If \(|a|=50\), then \(a=\) (A) \((24,32,30)\) (B) \((24,-32,30)\) (C) \((-24,32,30)\) (D) none of these

The vector \(\vec{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}\) lies in the plane of the vectors \(\vec{b}=\hat{i}+\hat{j}\) and \(\vec{c}=\hat{j}+\hat{k}\) and bisects the angle between \(\vec{b}\) and \(\vec{c}\). Then which one of the following gives possible values of \(\alpha\) and \(\beta ?\) (A) \(\alpha=2, \beta=2\) (B) \(\alpha=1, \beta=2\) (C) \(\alpha=2, \beta=1\) (D) \(\alpha=1, \beta=1\)

The vector differential operator DEL, written \(\nabla\), is defined by \(\nabla=\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j+\frac{\partial}{\partial z} k=i \frac{\partial}{\partial x}+j \frac{\partial}{\partial y}+k \frac{\partial}{\partial z}\), where \(\frac{\partial}{\partial x}\) rep- resents the derivative w.r.t. \(x\) regarding \(y\) and \(z\) as constant. Similarly, \(\frac{\partial}{\partial y}\) represents the derivative w.r.t. \(y\) regarding \(x\) and \(z\) as constant and \(\frac{\partial}{\partial z}\) represents the derivative w.r.t. \(z\) regarding \(x\) and \(y\) as constant. The operator \(\nabla\) is also known as nabla. Let \(\phi(x, y, z)\) be defined and differentiable at each point \((x, y, z)\) in a certain region of space. Then, the gradient of \(\phi\), written \(\nabla \phi\) or grad \(\phi\), is defined by $$ \nabla \phi=\left(\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j+\frac{\partial}{\partial z} k\right) \phi=\frac{\partial \phi}{\partial x} i+\frac{\partial \phi}{\partial y} j+\frac{\partial \phi}{\partial z} k $$ Let \(r\) be any vector such that \(r=x i+y j+z k\) \(\nabla r^{n}=\) (A) \(n r n-{ }^{1} r\) (B) \(n m-{ }^{2} r\) (C) \(\bar{n} \overline{r n} r\) (D) none of these

The sides of a parallelogram are \(2 i+4 j-5 k\) and \(i+2 j\) \(+3 k\). The unit vector parallel to one of the diagonals size is (A) \(\frac{1}{7}(3 i+6 j-2 k)\) (B) \(\frac{1}{7}(3 i-6 j-2 k)\) (C) \(\frac{1}{7}(-3 i+6 j-2 k)\) (D) \(\frac{1}{7}(3 i+6 j+2 k)\)

If \(A B=3 i+j-k\) and \(A C=i-j+3 k .\) If the point \(P\) on the line segment \(B C\) is equidistant from \(A B\) and \(A C\), then \(A P\) is (A) \(2 i-k\) (B) \(i-2 \mathrm{k}\) (C) \(2 i+k\) (D) none of these
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