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) \(\frac{1}{√69}(\hat{-i}-2\hat{j}+8\hat{k})\)
Match the LIST-I with LIST-II
LIST-I (Expressions) | LIST-II (Values) | ||
---|---|---|---|
A. | \( i^{49} \) | I. | 1 |
B. | \( i^{38} \) | II. | \(-i\) |
C. | \( i^{103} \) | III. | \(i\) |
D. | \( i^{92} \) | IV. | \(-1\) |
Choose the correct answer from the options given below:
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: