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.

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.

• 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 Feb 21, 2023 at 23:30
• You're welcome. My pleasure. Feb 21, 2023 at 23:33
• @Sunshine Please check my updated solution showing all $4$ solutions and which ones satisfy the acute angle with $\vec{c}$ condition. Feb 21, 2023 at 23:40
• Omg, thank You very much <3 I really appreciate Your help Feb 22, 2023 at 14:17