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

Consider the vectors $\vec{a}=3 \hat{i}+5 \hat{j}+2 \hat{k}$, $\vec{b}=2 \hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{c}=-5 \hat{i}-2 \hat{j}+3 \hat{k}$. If $l, m$ and $n$ are the lengths of the projections of $\vec{a}$ on $\vec{b}$, $\vec{b}$ on $\vec{c}$ and $\vec{c}$ on $\vec{a}$ respectively, then:

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 $\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$ and $|\vec{a} \times \vec{b}| = |\vec{a} \cdot \vec{b}|$, then $\theta$ is equal to

If $|a \times b|^2 + |a \cdot b|^2 = 36$ and $|a| = 3$,then $|b|$ is equal to

If the angle between the vectors $a$ and $b$ is $\theta$ and $a \cdot b = \cos \theta$,then the true statement is: