We are given the vector equations: \[ \alpha \overrightarrow{d} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \] \[ \beta \overrightarrow{a} = \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d} \] Step 1: Add the two given equations
Adding both equations, we get: \[ \alpha \overrightarrow{d} + \beta \overrightarrow{a} = (\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}) + (\overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d}) \] \[ \alpha \overrightarrow{d} + \beta \overrightarrow{a} = \overrightarrow{a} + \overrightarrow{d} + 2(\overrightarrow{b} + \overrightarrow{c}) \] Step 2: Rearrange and isolate vectors
Rearranging, we have: \[ (\alpha - 1) \overrightarrow{d} + (\beta - 1) \overrightarrow{a} = 2(\overrightarrow{b} + \overrightarrow{c}) \] Since \(\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}, \overrightarrow{d}\) are non-coplanar vectors, this equation implies the right side must be expressible as a linear combination of \(\overrightarrow{a}\) and \(\overrightarrow{d}\).
Step 3: Use non-coplanarity to find the sum
For this to hold, the only possibility is: \[ \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d} = \vec{0} \] Therefore, the magnitude of the sum is: \[ |\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} + \overrightarrow{d}| = 0 \]
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |
Let ABCD be a parallelogram and $ 2\bar{i} + \bar{j} $, $ 4\bar{i} + 5\bar{j} + 4\bar{k} $ and $ -\bar{i} - 4\bar{j} - 3\bar{k} $ be the position vectors of the vertices A, B, D respectively. Then the position vector of one of the points of trisection of the diagonal AC is