If ABCD is a parallelogram, vector AB = 2i + 4j - 5k and vector AD = i + 2j + 3k, then the unit vector in the direction of BD is
\(\frac{1}{√69}(\hat{i}+2\hat{j}-8\hat{k})\)
\(\frac{1}{69}(\hat{i}+2\hat{j}-8\hat{k})\)
\(\frac{1}{√69}(\hat{-i}-2\hat{j}+8\hat{k})\)
\(\frac{1}{69}(\hat{-i}-2\hat{j}+8\hat{k})\)
The correct answer is option C) \(\frac{1}{√69}(\hat{-i}-2\hat{j}+8\hat{k})\)
The respective values of \( |\vec{a}| \) and} \( |\vec{b}| \), if given \[ (\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) = 512 \quad \text{and} \quad |\vec{a}| = 3 |\vec{b}|, \] are:
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: