The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of vectors $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}+3 \hat{k}$ is equal to $1$. Find the value of $\lambda$.

Let $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} - 2\hat{k}$ be three vectors. $A$ vector of the type $\vec{b} + \lambda \vec{c}$ for some scalar $\lambda$,whose projection on $\vec{a}$ is of magnitude $\sqrt{\frac{2}{3}}$ is

Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be four vectors such that $\vec{a}$ is perpendicular only to $\vec{c}$. If the vector $\vec{b}$ is parallel to $(\vec{c}-\vec{d})$, then $\vec{c}$ is equal to:

If $\overrightarrow{a}=\hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}-3 \hat{j}+\hat{k}$ and the orthogonal projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\frac{4}{3}(\hat{i}-\hat{j}-\hat{k})$,then $\lambda$ is equal to

If $3\vec{a} - 5\vec{b}$ and $2\vec{a} + \vec{b}$ are perpendicular to each other,and $\vec{a} + 4\vec{b}$ and $-\vec{a} + \vec{b}$ are also perpendicular to each other,and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,then find $\cos \theta$.

If the vectors $a\,i - 2j + 3k$ and $3i + 6j - 5k$ are perpendicular to each other,then $a$ is given by