Question

Find the vector of magnitude 3√2 which lies in ZX plane and is at right angle to the vector 2i ⃗+j ⃗+2k ⃗ .

Answer 1

Question #43541, Math, Vector Calculus

Find the vector of magnitude $3\sqrt{2}$ which lies in zx plane and is at right angle to the vector $2\vec{i} + \vec{j} + 2\vec{k}$.

Solution.

We must find unknowing vector, which lies in ZX plane.

Let $\vec{a}$ is the unknown vector and this vector lies in ZX plane. It can be written as:

$$\vec{a} = x \cdot \vec{i} + 0 \cdot \vec{j} + z \cdot \vec{k}$$

where $x$ and $z$ some numbers. The y-coordinate of this vector is 0 because $\vec{a}$ lies in ZX plane.

Vector $\vec{a}$ is at right angle to the vector $2\vec{i} + \vec{j} + 2\vec{k}$, this means that

$$\vec{a} \cdot (2\vec{i} + \vec{j} + 2\vec{k}) = 0$$

We know that $|\vec{a}| = 3\sqrt{2}$ then we can write the system of equations:

$$\begin{array}{l} \left\{ \begin{array}{l} |\vec{a}| = 3\sqrt{2}, \\ \vec{a} \cdot (2\vec{i} + \vec{j} + 2\vec{k}) = 0; \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \sqrt{x^2 + z^2} = 3\sqrt{2}, \\ x \cdot 2 + 0 \cdot 1 + z \cdot 2 = 0; \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x^2 + z^2 = 18, \\ 2x + 2z = 0; \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} x^2 + z^2 = 18, \\ x + z = 0; \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x^2 + z^2 = 18, \\ z = -x; \end{array} \right. \Rightarrow x^2 + (-x)^2 = 18, \Rightarrow 2x^2 = 18, \Rightarrow \\ \Rightarrow x^2 = 9 \end{array}$$ $$\begin{array}{l} x^2 = 9 \Rightarrow \left\{ \begin{array}{l} x = 3, z = -3, \\ x = -3, z = 3; \end{array} \right. \\ \end{array}$$

check out these answers:

a) $\begin{cases}

\sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}, \\

(3\vec{i} - 3\vec{k}) \cdot (2\vec{i} + \vec{j} + 2\vec{k}) = 3 \cdot 2 - 3 \cdot 2 = 6 - 6 = 0

\end{cases}$

$\Rightarrow$ it's Ok

b) $\begin{cases}

\sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}, \\

(-3\vec{i} + 3\vec{k}) \cdot (2\vec{i} + \vec{j} + 2\vec{k}) = -3 \cdot 2 + 3 \cdot 2 = -6 + 6 = 0

\end{cases}$

$\Rightarrow$ it's Ok

Answer:

$$\vec{a} = \pm 3\vec{i} \mp 3\vec{k} = \begin{array}{c} 3\vec{i} - 3\vec{k} \\ -3\vec{i} + 3\vec{k} \end{array}$$

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