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The vectors $\vec a$ = (1, 1, 1), $\vec b$ = (−1, 2, 1) and $\vec c$ = (13, 2, 0) are given. Find the vector $\vec d$ whose modulus (intensity) is √6, which is perpendicular to the vector $\vec a$, forms an acute angle with the vector $\vec c$, and the surface of the parallelogram constructed over vectors $\vec b$ and $\vec d$ is 2√5.

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Let $\vec{d} = (d_1, d_2, d_3) $

Since $ | \vec{d} | = \sqrt{6} $ , then

$ d_1^2 + d_2^2 + d_3^2 = 6 $

Since $\vec{d}$ is perpendicular to $\vec{a}$ then

$ \vec{a} \cdot \vec{d} = 0 $

This translates into $ d_1 + d_2 + d_3 = 0 $

From here it follows that $d_3 = -d_1 - d_2 $

Using this in the expression for the squared length of $\vec{d}$ gives

$ 6 = d_1^2 + d_2^2 + (-d_1 - d_2)^2 = 2 (d_1^2 + d_2^2 + d_1 d_2 ) $

which reduces to

$ d_1^2 + d_2^2 + d_1 d_2 = 3 $

The area of the parallelogram constructed over $\vec{b}$ and $\vec{d} $ is given by

$ \text{Area} = 2 \sqrt{5} = | \vec{b} \times \vec{d} | = | (-1, 2, 1) \times (d_1, d_2, d_3) | = | (2 d_3 - d_2 , d_1 + d_3 , -d_2 - 2 d_1) |$

Squaring,

$20 = (2 d_3 - d_2)^2 + (d_1 + d_3)^2 + (d_2 + 2 d_1)^2$

Using $d_3 = -d_1 - d_2$, we get

$20 = (-2 d_1 -3 d_2) + d_2^2 + (d_2 + 2 d_1)^2 = 8 d_1^2 +11 d_2^2 + 16 d_1 d_2 $

Now we have two equation in $d_1$ and $d_2$. From the first one we have

$ d_1 d_2 = 3 - d_1^2 - d_2^2 $

Using this into the second equation

$ 20 = 48 - 8 d_1^2 - 5 d_2^2 $

So that

$ 8 d_1^2 + 5 d_2^2 = 28 $

The solution of this ellipse equation is

$ d_1 = \sqrt{ \dfrac{28}{8} } \cos \phi $

$ d_2 = \sqrt{ \dfrac{28}{5} } \sin \phi $

But $ \phi $ cannot take any value, because we must have

$ d_1 d_2 = 3 - d_1^2 - d_2^2 $

i.e.

$ \dfrac{28}{\sqrt{40}} \cos \phi \sin \phi = 3 - \dfrac{28}{8} \cos^2 \phi - \dfrac{28}{5} \sin^2 \phi $

Divide through by $28$

$ \dfrac{1}{2 \sqrt{40}} \sin(2 \phi) + \left(\dfrac{1}{16} - \dfrac{1}{10} \right) \cos(2 \phi) = \dfrac{3}{28} - \left(\dfrac{1}{16} + \dfrac{1}{10} \right) $

Solving this trigonometric equation is straightforward, and there are four solutions for $\phi \in [0, 2 \pi) $

Corresponding to each solution $\phi$ we can find $d_1, d_2, d_3$ and then check if $ \vec{c} \cdot \vec{d} \gt 0 $ as is required.

The four solutions are

$ \vec{d} = (1, -2, 1) $

$ \vec{d} = (-1, 2, -1) $

$ \vec{d} = (\dfrac{13}{7} , - \dfrac{2}{7} , -\dfrac{11}{7} )$

$ \vec{d} = (- \dfrac{13}{7}, \dfrac{2}{7} , \dfrac{11}{7} ) $

The corresponding dot products with $\vec{c}=(13,2,0)$ are, respectively, $ 9 , -9, \dfrac{ 165}{7} , - \dfrac{165}{7} $

Therefore, the two possible solutions are the first and the third.

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  • $\begingroup$ Thank You very much. And honestly, I was hoping there was some other easier solution :'( Thank You one more time, You helped me a lot <3 $\endgroup$
    – Sunshine
    Feb 21, 2023 at 23:30
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    $\begingroup$ You're welcome. My pleasure. $\endgroup$
    – of course
    Feb 21, 2023 at 23:33
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    $\begingroup$ @Sunshine Please check my updated solution showing all $4$ solutions and which ones satisfy the acute angle with $\vec{c}$ condition. $\endgroup$
    – of course
    Feb 21, 2023 at 23:40
  • $\begingroup$ Omg, thank You very much <3 I really appreciate Your help $\endgroup$
    – Sunshine
    Feb 22, 2023 at 14:17

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