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Let [tex] \rm\hat{i}, \hat{j} \: and \: \hat{k} [/tex] be the unit vector along the three positive coordinate axes. Let [tex] \rm\overset{ \to}a = 3 \hat{i} + \hat{j} - \hat{k} ,\\ \rm \overset{ \to}b = \hat{i} + b_{2} \hat{j} + b_3\hat{k}, \: \: \: \: \: \: \: b_{2}, b_{3} \in \mathbb{R}, \\ \rm\overset{ \to}c= c_{1} \hat{i} + c_{2} \hat{j} + c_3\hat{k}, \: \: \: \: \:c_{1},c_{2}, c_{3} \in \mathbb{R}[/tex]
be three vectors such that [tex] \rm b_2b_3 > 0, \overset{ \to}a \cdot\overset{ \to}b = 0[/tex] and [tex] \left(\begin{gathered} 0& \rm - c_{3} & \rm c_{2} \\ \rm c_{3}& \rm 0 & \rm - c_{1} \\ \rm - c_{2}& \rm c_{1} & \rm 0 \end{gathered} \right) \left(\begin{gathered} 1 \\ \rm b_{2} \\ \rm b_3\end{gathered} \right) = \left(\begin{gathered} \rm3 - c_{1} \\ \rm 1 - c_{2} \\ \rm - 1 - c_3\end{gathered} \right)[/tex]
Then which of the following is true

[tex] \rm(1) \: \overset{ \to}a \cdot\overset{ \to}c = 0 \\ \rm(2) \: \overset{ \to}b \cdot\overset{ \to}c = 0 \\ (3) \: \overset{ \to}{| \rm b|} > \sqrt{10} \\(4) \: \overset{ \to}{| \rm c|} \leq \sqrt{11}[/tex]​

Let [tex] \rm\hat{i}, \hat{j} \: and \: \hat{k} [/tex] be the unit vector along the three positive coordinate axes. Let [tex] \rm\overset{ \to}a = 3 \hat{i} + \ class=

Answer :

LammettHash

Computing the matrix product in the last condition gives a system of linear equations in the components of [tex]\vec c[/tex],

[tex]\begin{cases} c_1 + b_3c_2 - b_2c_3 = 3 \\ -b_3c_1 + c_2 + c_3 = 1 \\ b_2c_1 - c_2 + c_3 = -1 \end{cases}[/tex]

and is easily solvable to get

[tex]c_1 = \dfrac{6-2b_3}{2+{b_2}^2+{b_3}^2}, c_2 = \dfrac{2+3b_3}{2+{b_2}^2+{b_3}^2}, c_3 = \dfrac{9-3b_3}{2+{b_2}^2+{b_3}^2}[/tex]

or, using the condition [tex]\vec a\cdot\vec b = 3 + b_2 - b_3 = 0[/tex],

[tex]c_1 = -\dfrac{2b_2}{11 + 6b_2 + 2{b_2}^2}, c_2 = \dfrac{11+3b_2}{11+6b_2+2{b_2}^2}, c_3 = -\dfrac{3b_2}{11+6b_2+2{b_2}^2}[/tex]

• (1) is false, since

[tex]\vec a \cdot \vec c = \dfrac{11}{2 + {b_2}^2 + {b_3}^2} \neq 0[/tex]

because [tex]{b_2}^2+{b_3}^2\ge0 \implies 0<\vec a\cdot\vec c\le\frac{11}2[/tex].

• (2) is true.

[tex]\vec b\cdot \vec c = \dfrac{(3 + b_2 - b_3) (2 + 3b_3)}{2 + {b_2}^2 + {b_3}^2} = \dfrac{(\vec a\cdot\vec b)(2+3b_3)}{2+{b_2}^2+{b_3}^2} = 0[/tex]

• (3) is false. The magnitude of [tex]\vec b[/tex] could be smaller than √10.

Since [tex]\vec a\cdot\vec b=0[/tex], we have

[tex]\|\vec b\| = \sqrt{1 + {b_2}^2 + (3+b_2)^2} = \sqrt{10 + 6b_2 + 2{b_2}^2}[/tex]

and

[tex]2{b_2}^2 + 6b_2 = 2\left({b_2}^2 + 3b_2 + \dfrac94\right) - \dfrac{18}4 = 2\left(b_2 + \dfrac32\right)^2 - \dfrac92 \ge - \dfrac92[/tex]

which is to say, [tex]\|\vec b\| \ge \sqrt{10-\frac92} = \sqrt{\frac{11}2}[/tex].

• (4) is true.

We have

[tex]\|\vec c\| = \dfrac{\sqrt{11} \sqrt{11 - 6b_3 + 2{b_3}^2}}{2 + {b_2}^2 + {b_3}^2}[/tex]

In terms of [tex]b_2[/tex] alone,

[tex]\|\vec c\| = \dfrac{\sqrt{11}\sqrt{11 + 6b_2 + 2{b_2}^2}}{11+6b_2+2{b_2}^2} = \dfrac{\sqrt{11}}{\sqrt{11+6b_2+2{b_2}^2}}[/tex]

Now,

[tex]11 + 6b_2 + 2{b_2}^2 = 2\left(b_2+\dfrac32\right)^2 + \dfrac{13}2 \ge \dfrac{13}2[/tex]

which means

[tex]\|\vec c\| \le \dfrac{\sqrt{11}}{\sqrt{11 + \frac{13}2}} \le \sqrt{11}[/tex]

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