Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
Answer:
[tex]\|a\| = 5\sqrt{13}[/tex].
[tex]\|b\| = 3\sqrt{29}[/tex].
Step-by-step explanation:
Let [tex]m[/tex],[tex]n[/tex], and [tex]k[/tex] be scalars such that:
[tex]\displaystyle a = m\, (2\, \vec{i} + 3\, \vec{j}) = m\, \begin{bmatrix}2 \\ 3\end{bmatrix}[/tex].
[tex]\displaystyle b = n\, (2\, \vec{i} + 5\, \vec{j}) = n\, \begin{bmatrix}2 \\ 5\end{bmatrix}[/tex].
[tex]\displaystyle (a + b) = k\, (8\, \vec{i} + 15\, \vec{j}) = k\, \begin{bmatrix}8 \\ 15\end{bmatrix}[/tex].
The question states that [tex]\| a + b \| = 34[/tex]. In other words:
[tex]k\, \sqrt{8^{2} + 15^{2}} = 34[/tex].
[tex]k^{2} \, (8^{2} + 15^{2}) = 34^{2}[/tex].
[tex]289\, k^{2} = 34^{2}[/tex].
Make use of the fact that [tex]289 = 17^{2}[/tex] whereas [tex]34 = 2 \times 17[/tex].
[tex]\begin{aligned}17^{2}\, k^{2} &= 34^{2}\\ &= (2 \times 17)^{2} \\ &= 2^{2} \times 17^{2} \end{aligned}[/tex].
[tex]k^{2} = 2^{2}[/tex].
The question also states that the scalar multiple here is positive. Hence, [tex]k = 2[/tex].
Therefore:
[tex]\begin{aligned} (a + b) &= k\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 2\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 16\, \vec{i} + 30\, \vec{j}\\ &= \begin{bmatrix}16 \\ 30 \end{bmatrix}\end{aligned}[/tex].
[tex](a + b)[/tex] could also be expressed in terms of [tex]m[/tex] and [tex]n[/tex]:
[tex]\begin{aligned} a + b &= m\, (2\, \vec{i} + 3\, \vec{j}) + n\, (2\, \vec{i} + 5\, \vec{j}) \\ &= (2\, m + 2\, n) \, \vec{i} + (3\, m + 5\, n) \, \vec{j} \end{aligned}[/tex].
[tex]\begin{aligned} a + b &= m\, \begin{bmatrix}2\\ 3 \end{bmatrix} + n\, \begin{bmatrix} 2\\ 5 \end{bmatrix} \\ &= \begin{bmatrix}2\, m + 2\, n \\ 3\, m + 5\, n\end{bmatrix}\end{aligned}[/tex].
Equate the two expressions and solve for [tex]m[/tex] and [tex]n[/tex]:
[tex]\begin{cases}2\, m + 2\, n = 16 \\ 3\, m + 5\, n = 30\end{cases}[/tex].
[tex]\begin{cases}m = 5 \\ n = 3\end{cases}[/tex].
Hence:
[tex]\begin{aligned} \| a \| &= \| m\, (2\, \vec{i} + 3\, \vec{j})\| \\ &= m\, \| (2\, \vec{i} + 3\, \vec{j}) \| \\ &= 5\, \sqrt{2^{2} + 3^{2}} = 5 \sqrt{13}\end{aligned}[/tex].
[tex]\begin{aligned} \| b \| &= \| n\, (2\, \vec{i} + 5\, \vec{j})\| \\ &= n\, \| (2\, \vec{i} + 5\, \vec{j}) \| \\ &= 3\, \sqrt{2^{2} + 5^{2}} = 3 \sqrt{29}\end{aligned}[/tex].
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.