Question

Let $\vec a = \hat i - \hat j$, $\vec b = \hat i + \hat j + \hat k$ and $\vec c$ be a vector such that $\vec a \times \vec c + \vec b = \vec 0$ and $\vec a \cdot \vec c = 4$, then ${\left| {\vec c} \right|^2}$ is equal to

Answer 1

19/2

Answer 2

Given the vector equation $\vec a \times \vec c + \vec b = \vec 0$, we can rewrite it as $\vec a \times \vec c = -\vec b$. Taking the magnitude squared of both sides, we get $|\vec a \times \vec c|^2 = |-\vec b|^2 = |\vec b|^2$. Using the vector identity $|\vec a \times \vec c|^2 = |\vec a|^2 |\vec c|^2 - (\vec a \cdot \vec c)^2$, we substitute this into the equation: $|\vec a|^2 |\vec c|^2 - (\vec a \cdot \vec c)^2 = |\vec b|^2$.

We are given $\vec a = \hat i - \hat j$, so its squared magnitude is $|\vec a|^2 = (1)^2 + (-1)^2 + (0)^2 = 1 + 1 = 2$. We are given $\vec b = \hat i + \hat j + \hat k$, so its squared magnitude is $|\vec b|^2 = (1)^2 + (1)^2 + (1)^2 = 1 + 1 + 1 = 3$. We are also given that $\vec a \cdot \vec c = 4$.

Substitute these values into the equation: $2 |\vec c|^2 - (4)^2 = 3$ $2 |\vec c|^2 - 16 = 3$ $2 |\vec c|^2 = 19$ $|\vec c|^2 = \frac{19}{2}$.