\(\underline{a}\) and \(\underline{b}\) are non coplanar and \(5 \underline{u}-3 \underline{\mathrm{v}}=\underline{\mathrm{w}} .\) if \(\underline{u}=\mathrm{ma}+2 \mathrm{nb}\) and \(\underline{\mathrm{v}}=-2 \mathrm{na}+3 \mathrm{mb}\) and \(\underline{\mathrm{w}}=4 \underline{\mathrm{a}}-2 \underline{\mathrm{b}} .\) find \(\mathrm{m}=\underline{\mathrm{n}}=\) (a) \((1 / 2),(1 / 4)\) (b) \((1 / 2),(1 / 3)\) (c) \((1 / 3),(1 / 4)\) (d) \(-(1 / 2),-(1 / 4)\)

Short Answer

Expert verified
The short answer is: \(m = \frac{10}{19}\), \(n = \frac{8}{57}\). However, this option is not provided in the available choices, which indicates an error in the problem or the given options.

Step by step solution

01

Substitute the expressions for u, v and w in the given equation

Given: \(5\underline{u} - 3\underline{v} = \underline{w}\) Now, we will substitute the given expressions of \(\underline{u}\), \(\underline{v}\), and \(\underline{w}\) into the equation: \(5(m\underline{a} + 2n\underline{b}) - 3(-2n\underline{a} + 3m\underline{b}) = 4\underline{a} - 2\underline{b}\)
02

Distribute the scalars and reorganize the terms

Next, we distribute the scalars to each of the vectors and rearrange the terms: \(5m\underline{a} + 10n\underline{b} + 6n\underline{a} - 9m\underline{b} = 4\underline{a} - 2\underline{b}\) Now group \(\underline{a}\) and \(\underline{b}\) terms: \((5m + 6n)\underline{a} + (10n - 9m)\underline{b} = 4\underline{a} - 2\underline{b}\)
03

Set up the equations for m and n

Since the vectors \(\underline{a}\) and \(\underline{b}\) are non-coplanar, this can be treated as a system of two equations in the coefficients \(m\) and \(n\). \begin{align*} 5m + 6n &= 4 \\ 10n - 9m &= -2 \end{align*}
04

Solve the system of equations for m and n

To solve for m and n, we can multiply the first equation by 2 to make the coefficients of m match, and then subtract the second equation from the result: \begin{align*} &2(5m + 6n) = 2(4) \\ &10m + 12n = 8 \\ \\ &10m + 12n - (10n - 9m) = 8 - (-2) \\ &19m = 10 \\ &m = \frac{10}{19} \end{align*} Now, substitute \(m\) into the first equation to solve for \(n\): \[ 5\left(\frac{10}{19}\right) + 6n = 4 \] Divide through by 5 to simplify: \[ \frac{10}{19} + \frac{6}{5}n = \frac{4}{5} \] Now isolate \(n\): \[ \frac{6}{5}n = \frac{4}{5} - \frac{10}{19} \] Find a common denominator (19) and subtract: \[ \frac{6}{5}n = \frac{16}{95} \] Now divide through by \(\frac{6}{5}\): \[ n = \frac{8}{57} \] Therefore, the values of m and n are: \[ m = \frac{10}{19}, \quad n = \frac{8}{57} \] This option is not provided in the multiple-choice answers. There seems to be an error in the problem or the given options.

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Most popular questions from this chapter

if \(\underline{a}+m \underline{b}+3 \underline{c},-2 \underline{a}+3 \underline{b}-4 \underline{c}\) and \(\underline{a}-3 \underline{b}-5 \underline{c}\) are coplanar. \(\mathrm{m}=\) (a) 2 (b) \(-1\) (c) 1 (d) \(-(9 / 7)\)

\(\underline{a}=(1 / \sqrt{1} 0)(3 \underline{i}+\underline{k}) \underline{b}=(1 / 7)(2 \underline{i}+3 j-6 \underline{k})\) evaluate \((2 \underline{\mathrm{a}}-\underline{\mathrm{b}}) \cdot[(\underline{\mathrm{a}} \times \underline{\mathrm{b}}) \times(\underline{\mathrm{a}}+2 \underline{\mathrm{b}})]\) (a) \(-3\) (b) 5 (c) 3 (d) \(-5\)

if \(|\underline{\mathrm{A}}|=3,|\underline{\mathrm{B}}|=4,|\underline{\mathrm{C}}|=5, \underline{\mathrm{A}} \perp(\underline{\mathrm{B}}+\underline{\mathrm{C}}), \underline{\mathrm{B}} \perp(\underline{\mathrm{C}}+\underline{\mathrm{A}})\), \(\underline{\mathrm{C}} \perp(\underline{\mathrm{A}}+\underline{\mathrm{B}})\) then \(|\underline{\mathrm{A}}+\underline{\mathrm{B}}+\underline{\mathrm{C}}|=\) (a) \(5 \sqrt{2}\) (b) \(7 \sqrt{2}\) (c) \(\sqrt{2}\) (d) \(3 \sqrt{2}\)

if \(\underline{\mathrm{OA}}=\underline{\mathrm{a}}, \underline{\mathrm{OB}}=\underline{\mathrm{b}}, \underline{\mathrm{OC}}=2 \underline{\mathrm{a}}+3 \underline{\mathrm{b}}, \underline{\mathrm{OD}}=\underline{\mathrm{a}}-2 \underline{\mathrm{b}}\). Length of \(\underline{\mathrm{OA}}\) is three times length of \(\underline{\mathrm{OB}} \& \mathrm{OA}\) is perpendicular to \(\underline{\mathrm{DB}}\) then \((\underline{\mathrm{BD}} \times \underline{\mathrm{AC}}) \cdot(\underline{\mathrm{OD}} \times \underline{\mathrm{OC}})\) is (a) \(7|\underline{\mathrm{a}} \times \underline{\mathrm{b}}|^{2}\) (b) 0 (c) \(7|\underline{a} \times \underline{b}|\) (d) \(42|\underline{\mathrm{a}} \times \underline{\mathrm{b}}|^{2}\)

if \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) and \(\mathrm{p}, \mathrm{q}, \mathrm{r}\) are reciprocal system of vectors then \(a \times p+b \times q+c \times r\) is equal to (a) \([\mathrm{a} \mathrm{b} \mathrm{c}]\) (b) \(3(a+b+c)\) (c) 0 (d) None of these.

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