A student writes 2(1, 0) + 4(-1, 0) =(-2,0) so (1, 0), (-1, 0) span R^2. What is wrong with this conclusion?
It is claimed that \{(1, 0, 0), (0, 1, 0), (0, 0, 0)\} is a set of vectors spanning R^3. Explain why it is not possible for these vectors to span R^3.
Describe the set of vectors spanned by (0, 1). Say why this is the same set as that spanned by (0, -1).
In R^3, the vector \vec{i} = (1, 0, 0), spans a set. Describe the set spanned by this vector. Name two other vectors that would also span the same set.
It is proposed that the set \{(0, 0), (1, 0)\} could be used to span R^2. Explain why this is not possible.
The following is a spanning set for R^2.
\{(-1, 2), (2, -4), (-1, 1), (-3, 6), (1, 0)\}
Remove three of the vectors and write down a spanning set that can be used to span R^2.
Simplify each of the following linear combinations and write your answer in component form:
\vec{a} = \vec{i} - 2\vec{j}, \vec{b} = \vec{j} - 3\vec{k}, and \vec{c}= \vec{i} - 3\vec{j} + 2\vec{k}
2(2\vec{a} - 3\vec{b} + \vec{c}) - 4(-\vec{a} + \vec{b} - \vec{c}) + (\vec{a} -\vec{c})
Simplify each of the following linear combinations and write your answer in component form:
\vec{a} = \vec{i} - 2\vec{j}, \vec{b} = \vec{j} - 3\vec{k}, and \vec{c}= \vec{i} - 3\vec{j} + 2\vec{k}
\displaystyle{\frac{1}{2}}(2\vec{a} - 4\vec{b} - 8\vec{c}) - \displaystyle{\frac{1}{3}}(3\vec{a} - 6\vec{b} + 9\vec{c})
Name two sets of vectors that could be used to span the xy-plane in R^3. Show how the vectors (-1, 2, 0) and (3, 4, 0) could each be written as a linear combination of the vectors you have chosen.
a) The set of vectors {(1, 0, 0), (0, 1, 0)} spans a set in \mathbb{R}^3. Describe this set.
b) Write the vector (-2, 4, 0) as a linear combination of these vectors.
c) Explain why it is not possible to write (3, 5, 8) as a linear combination of these vectors.
d) If the vector (1, 1, 0) were added to this set, what would these three vectors span in \mathbb{R}^3?
Solve for a, b, and c in the following equation:
2(a, 3, c) + 3(c, 7, c) = (5, b + c, 15)
Write the vector (-10, -34) as a linear combination of the vectors (-1, 3) and (1, 5).
Find a and b which will satisfy to give you equation of a line give two position vectors \vec{u} and \vec{b}: (x, y) = a\vec{u} + b\vec{v}.
Given two position vectors {(2, -1), (-1, 1)}
Write each of the following vectors as a linear combination of the set given in part a: (2, -3), (124, -5), and (4, -11).
Show that the vectors (-1,2, 3), (4, 1, -2), and (-14, -1, 16) do not lie on the same plane.
Show that the vectors (-1, 3, 4), (0, -1, 1), and (-3, 14, 7) lie on the same plane, and show how one of the vectors can be written as a linear combination of the other two.
Determine the value for x such that the points A(-1, 3, 4), B(-2, 3, -1), and C(-5, 6, x) all lie on a plane that contains the origin.
The vectors \vec{a} and \vec{b} span \mathbb{R}^2. What values of m and n will make the following statements true: (m - 2)\vec{a} = (n + 3)\vec{b}? Explain your reasoning.
The vectors (4, 1, 7), (-1, 1, 6) and (p, q, 5) are coplanar. Determine three sets of values for p and q for which this is true?
The vectors \vec{a} and \vec{b} span \mathbb{R}^2. For what values of m is it true that (m^2 + 2m - 3)\vec{a} + (m^2 + m - 6)\vec{b} = \vec{0}? Explain your reasoning.
Introduction to Linear Combination
\vec{c} = k\vec{a} + l\vec{b}
That means \vec a, \vec b, \vec c are coplaner
Geometrically example of linear combination of vectors ex
ex Three vectors \vec u, \vec v, and \vec x have magnitudes |\vec u| = 10, |\vec v| = 15, and |\vec x| = 24. If x lies between \vec u and \vec v in the same plane, making an angle of 20 with \vec u and 30 with \vec v, express \vec x as a linear combination of u and v.
Therefore, \vec x = 1.57 \vec u + 0.71 \vec v
Conceptual explanation of Spanning Sets
Finding coefficients to represent a vector in 2D as linear combination of two other given vectors
ex \vec x = (4, 36) as linear combination of
a) {(-1, 2), (2, 1)}
b) {(1, 0), (-2, 1)}
a) Therefore, \vec x = \dfrac{68}{5}(-1, 2) + \dfrac{44}{5}(2, 1)
b) Therefore, \vec x = 76(1, 0) + 36(-2, 1)
How to prove if two vectors span 2D
Therefore, \vec a, \vec b spans \mathbb{R} ^2
Why two parallel vector does not span vectors in 2D with an example
ex Span \vec a = (1, 3), \vec b = (2, 6)
Therefore, there is no solution.
Conceptual explanation of why we need 3 non coplanar vectors to span 3D
3 linearly independent vectors will span \mathbb{R}^3.
\vec a, \vec b, \vec c
\vec a = m\vec b + n\vec c *
No m, n can be found so that * is true.
Selecting a Linear combination Strategy to determine if vectors lie on the same plane
ex
a) Given the two vectors \vec a = (1, 2, 1) and \vec b = (3, 1, 1), does the vector \vec c = (-9, 3, 1) lie on the plane determine by \vec a and \vec b? Explain.
Therefore, the vector \vec c does not lie on the same plane determined by \vec a and \vec b.
b) Does the vector (-1, 3, 1) lie in the plane determined by the first two vectors
Therefore, (-1, 3, 1) is coplanar with \vec a, \vec b