A) 15
B) \[\sqrt{113}\]
C) \[\sqrt{593}\]
D) \[\sqrt{369}\]
Correct Answer: C
Solution :
Given that, \[\left| a \right|=2\sqrt{2},\left| b \right|=3\] The longer vector is 5a + 2b + a ? 3b = 6a ? b Length of one diagonal = |6a ? b| \[=\sqrt{36{{a}^{2}}+{{b}^{2}}-2\times 6|a|.|b|.\cos {{45}^{\circ }}}\] \[=\sqrt{36\times 8+9-12\times 2\sqrt{2}\times 3\times \frac{1}{\sqrt{2}}}\] \[=\sqrt{288+9-12\times 6}=\sqrt{225}=15\] Other diagonal is 4a + 5b. Its length is \[=\sqrt{{{\left( 4a \right)}^{2}}+{{\left( 5b \right)}^{2}}+2x|4a||5a|\cos {{45}^{\circ }}}\] \[=\sqrt{16\times 8+25\times 9+40\times 6}=\sqrt{593}\]
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