Let $$\vec{a}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \vec{b}=\hat{i}+\hat{k}$$ and $$\vec{c}=\hat{i}+2 \hat{j}-3 \hat{k}$$ be three given vectors. If $$\overrightarrow{\mathrm{r}}$$ is a vector such that $$\vec{r} \times \vec{a}=\vec{c} \times \vec{a}$$ and $$\vec{r} \cdot \vec{b}=0$$, then $$|\vec{r}|$$ is equal to :
Let $$\vec{a}=2 \hat{i}+\hat{j}+\hat{k}$$, and $$\vec{b}$$ and $$\vec{c}$$ be two nonzero vectors such that $$|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$$ and $$\vec{b} \cdot \vec{c}=0$$. Consider the following two statements:
(A) $$|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$$ for all $$\lambda \in \mathbb{R}$$.
(B) $$\vec{a}$$ and $$\vec{c}$$ are always parallel.
Then,
If $((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}$,
then $|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$ is equal to :
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