Animated Solution for Mathematics - Vector Algebra: If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=4\hat{i}+3\hat{j}+4\hat{k}$ and $\vec{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$ are linearly dependent vectors and $|\vec{c}|=\sqrt{3}$, then

Imagine standing in a three-dimensional space with two vectors, $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=4\hat{i}+3\hat{j}+4\hat{k}$. These vectors define a unique plane.

We introduce a third vector, $\vec{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$. The problem states that these three vectors are linearly dependent.

Geometrically, this implies that vector $\vec{c}$ lies perfectly flat within the plane defined by $\vec{a}$ and $\vec{b}$.

To translate this "trapped" state into mathematics, we use the scalar triple product. If three vectors are coplanar, the volume of the parallelepiped they form must be zero.

Mathematically, this volume is calculated using the determinant of the matrix formed by their components:

We expand this determinant along the first row:

Distributing the values, we observe that the terms involving $\alpha$ cancel out:

Simplifying this expression, we obtain:

We now turn to the second constraint: the magnitude of vector $\vec{c}$ is $\sqrt{3}$. The magnitude formula in 3D is given by:

Setting this equal to $\sqrt{3}$ and squaring both sides, we get:

Substituting our known value $\beta =1$ into the equation:

This simplifies to ${\alpha }^2=1$, which yields $\alpha =\pm 1$.

We have successfully navigated the constraints, finding $\beta =1$ and $\alpha =\pm 1$.