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The vector (a) is a multiple of the vector (2i +3j) and (b) is a multiple of (2i+5j) The sum (a+b) is a multiple of the vector (8i +15j). Given that /a+b/= 34 and the scaler multiple of (8i+15j) is positive, Find the magnitude of a and b.​

Sagot :

jacob193

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].

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