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

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Vector Algebra

Suppose that

p, q

and

r

are three non-coplanar vectors in

R^{3}

. Let the components of a vector

s

along

p, q

and

r

be 4, 3 and 5, respectively. If the components of this vector

s

along

(- p + q + r), (p - q + r)

and

(- p - q + r)

are

x, y

and

z

, respectively, then the value of

2 x + y + z

Answer: 9

We have

s = 4 p + 3 q + 5 r

. Also,

s = x (- p + q + r) + y (p - q + r) + z (- p - q + r)

. Expanding and equating coefficients:

Adding the first two equations:

- 2 z = 7

, so

z = - \frac{7}{2}

. From the third equation:

x + y = 5 - z = 5 + \frac{7}{2} = \frac{17}{2}

. From the first equation:

- x + y = 4 + z = 4 - \frac{7}{2} = \frac{1}{2}

. Solving:

2 y = \frac{17}{2} + \frac{1}{2} = 9

, so

y = \frac{9}{2}

and

x = 4

. Thus

2 x + y + z = 8 + \frac{9}{2} - \frac{7}{2} = 8 + 1 = 9

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Integer answer