Let $$\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$$. Then the square of the projection of $$\vec{a}$$ on $$\vec{b}$$ is:
Let $$\overrightarrow{\mathrm{a}}=6 \hat{i}+\hat{j}-\hat{k}$$ and $$\overrightarrow{\mathrm{b}}=\hat{i}+\hat{j}$$. If $$\overrightarrow{\mathrm{c}}$$ is a is vector such that $$|\overrightarrow{\mathrm{c}}| \geq 6, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=6|\overrightarrow{\mathrm{c}}|,|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2}$$ and the angle between $$\vec{a} \times \vec{b}$$ and $$\vec{c}$$ is $$60^{\circ}$$, then $$|(\vec{a} \times \vec{b}) \times \vec{c}|$$ is equal to:
If the area of the region $$\left\{(x, y): \frac{\mathrm{a}}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0<\mathrm{a}<1\right\}$$ is $$\left(\log _{\mathrm{e}} 2\right)-\frac{1}{7}$$ then the value of $$7 \mathrm{a}-3$$ is equal to :
Let $$\mathrm{P}(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{Q}(3,-3,1)$$ in the line $$\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}$$ and $$\mathrm{R}$$ be the point $$(2,5,-1)$$. If the area of the triangle $$\mathrm{PQR}$$ is $$\lambda$$ and $$\lambda^2=14 \mathrm{~K}$$, then $$\mathrm{K}$$ is equal to :