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JEE Advance - Mathematics (2015 - Paper 2 Offline - No. 3)

Suppose that $$\overrightarrow p ,\overrightarrow q $$ and $$\overrightarrow r $$ are three non-coplanar vectors in $${R^3}$$. Let the components of a vector $$\overrightarrow s $$ along $$\overrightarrow p ,$$ $$\overrightarrow q $$ and $$\overrightarrow r $$ be $$4, 3$$ and $$5,$$ respectively. If the components of this vector $$\overrightarrow s $$ along $$\left( { - \overrightarrow p + \overrightarrow q + \overrightarrow r } \right),\left( {\overrightarrow p - \overrightarrow q + \overrightarrow r } \right)$$ and $$\left( { - \overrightarrow p - \overrightarrow q + \overrightarrow r } \right)$$ are $$x, y$$ and $$z,$$ respectively, then the value of $$2x+y+z$$ is
Answer
9

Explanation

Here, $$\overrightarrow s = 4\overrightarrow p + 3\overrightarrow q + 5\overrightarrow r $$ ....... (i)

and $$\overrightarrow s = ( - \overrightarrow p + \overrightarrow q + \overrightarrow r )x + (\overrightarrow p - \overrightarrow q + \overrightarrow r )y + ( - \overrightarrow p - \overrightarrow q + \overrightarrow r )z$$ ...... (ii)

$$\therefore$$ $$4\overrightarrow p + 3\overrightarrow q + 5\overrightarrow r = \overrightarrow p ( - x + y - z) + \overrightarrow q (x - y - z) + \overrightarrow r (x + y + z)$$

On comparing both sides, we get

$$ - x + y - z = 4$$, $$x - y - z = 3$$ and $$x + y + z = 5$$

On solving above equations, we get

$$x = 4$$, $$y = {9 \over 2}$$, $$z = {{ - 7} \over 2}$$

$$\therefore$$ $$2x + y + z = 8 + {9 \over 2} - {7 \over 2} = 9$$

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