Q.If $(1, 2, 4)$ and $(2, -3\lambda, -3)$ are the initial and terminal points of the vector $\hat i+5\hat j-7\hat k$, then the value of $\lambda$ is equal to

Resolving in the plane. Let $\hat i,\hat j$ be unit vectors along the positive $x$- and $y$-axes. Every position vector in the plane is written uniquely as $\vec{OP}=x\hat i+y\hat j,\qquad |\vec{OP}|=\sqrt{x^2+y^2}.$ More generally, if $\vec a,\vec b$ are any two non-collinear vectors in a plane, every vector in that plane is a unique linear combination $\lambda\vec a+\mu\vec b$.

Resolving in space. With $\hat i,\hat j,\hat k$ along the positive $x,y,z$ axes, every position vector in space is uniquely $\vec{OP}=x\hat i+y\hat j+z\hat k,\qquad |\vec{OP}|=\sqrt{x^2+y^2+z^2}.$ Three non-coplanar vectors $\vec a,\vec b,\vec c$ likewise span all of space uniquely: any vector is $\lambda\vec a+\mu\vec b+\nu\vec c$. The vector joining $(x_1,y_1,z_1)$ to $(x_2,y_2,z_2)$ has components $(x_2-x_1)\hat i+(y_2-y_1)\hat j+(z_2-z_1)\hat k$.

Matrix form. $\vec A=a_1\hat i+a_2\hat j+a_3\hat k$ can be written as the column $\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$; addition and scalar multiplication of vectors then match ordinary matrix addition and scalar multiplication component-by-component. …