Vector Algebra
If the vectors $$2\hat{i}-3\hat{j}+6\hat{k}$$, $$\vec{b}$$ are collinear and $$|b|=14$$, then $$\vec{b}=$$
Show that the vectors $$2\hat {i}-3\hat {j}+4\hat {k}$$ and $$-4\hat {i}+6\hat {j}-8\hat {k}$$ are collinear
Let $$\vec {a}=2\hat {i}-3\hat {j}+4\hat {k}$$ and $$\vec {b}=-4\hat {i}+6\hat {j}-8\hat {k}$$
It is observed that $$\vec {b}=4\hat {i}+6\hat {j}-8\hat {k}=-2(2\hat {i}-3\hat {j}+4\hat {k})=-2\vec {a}$$
$$\therefore \vec {b}=\lambda \vec {a},$$ where $$\lambda =-2$$
Hence, the given vectors are collinear.
If the vectors $$2\hat{i}-3\hat{j}+6\hat{k}$$, $$\vec{b}$$ are collinear and $$|b|=14$$, then $$\vec{b}=$$
Given three non-zero, non-coplanar vectors $$\vec{a}, \vec{b}$$ and $$\vec{c}$$. $$ \vec{r_1} = p\vec{a} + q\vec{b} + \vec{c}$$ and $$\vec{r_2} = \vec{a} + p\vec{b} + q\vec{c}$$. If the vectors $$\vec{r_1} + 2\vec{r_2}$$ and $$2\vec{r_1}+ \vec{r_2}$$ are collinear, then $$(p, q)$$ is
If $$a = ( 1 , -1 )$$ and $$b = (- 2 , m )$$ are two collinear vectors then $$m$$ is equal to :
If the vector $$\vec {b}$$ is colllinear with the vector $$\vec {a}=(2\sqrt {2},-1,4)$$ and $$|\vec {b}|=10$$, then:
If $$\vec {a}$$ and $$\vec {b}$$ are two collinear vectors, then which of the following are incorrect?
$$\vec{b} \space and \space \vec{c}$$ are non-collinear if $$\vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} . \vec{b})\vec{b} = (4 - 2x - sin y) \vec{b} + (x^2 - 1)\vec{c} \space and \space (\vec{c} . \vec{c})\vec{a} = \vec{c}$$. Then
If vectors $$\vec{a}\space and \space \vec{b}$$ are non-collinear, then $$\frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|}$$ is
Given three vector $$\vec a, \vec b$$ and $$\vec c$$, two of which are non-collinear. Further if $$(\vec a + \vec b)$$ is collinear with $$\vec c, (\vec b + \vec c)$$ is collinear with $$\vec a, \mid \vec a \mid = \mid \vec b \mid = \mid \vec c \mid = \sqrt 2$$. Find the value of $$\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a.$$
The points with position vectors $$\vec {a} + \vec {b}, \vec {a} - \vec {b}$$ and $$\vec {a} + k\vec {b}$$ are collinear for all real values of $$k$$.
If $$\vec{\alpha} + \vec{\beta} + \vec{\gamma} = a \vec{\delta}$$ and $$\vec{\beta} + \vec{\gamma} + \delta = b \vec{\alpha}, \vec{\alpha}$$ and $$\vec{\delta}$$ are non-collinear, then $$\vec{\alpha} + \vec{\beta} + \vec{\gamma} + \vec{\delta}$$ equals