Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies the equation p × [(x - q) × p ] + q × [(x - r) × q ] + r × [(x - p) × r ] = 0 then x is given by

Let p, q, r be three mutually perpendicular vectors of the… — JEE Main 1997 Mathematics PYQ

Let

\vec{p}, \vec{q}, \vec{r}

be three mutually perpendicular vectors of the same magnitude. If a vector

\vec{x}

satisfies the equation

\vec{p} \times \left[(\vec{x} - \vec{q}) \times \vec{p}\right] + \vec{q} \times \left[(\vec{x} - \vec{r}) \times \vec{q}\right] + \vec{r} \times \left[(\vec{x} - \vec{p}) \times \vec{r}\right] = \vec{0}

then

\vec{x}

is given by

A.
$\frac{1}{2}(\vec{p} + \vec{q} - 2\vec{r})$
B.
$\frac{1}{2}(\vec{p} + \vec{q} + \vec{r})$
C.
$\frac{1}{3}(\vec{p} + \vec{q} + \vec{r})$
D.
$\frac{1}{3}(2\vec{p} + \vec{q} - \vec{r})$

Show answer and solution

Correct option: B

Solution

Using the vector triple product identity: a × [(b) × a] = b(a·a) - a(a·b), we expand each term.

p × [(x - q) × p] = (x - q)|p|² - p[p·(x - q)]

Similarly for the other two terms. Summing all three:

|p|²[(x-q) + (x-r) + (x-p)] - p[p·(x-q)] - q[q·(x-r)] - r[r·(x-p)] = 0

Since p, q, r are mutually perpendicular with the same magnitude c², we have p·q = p·r = q·r = 0 and |p|² = |q|² = |r|² = c².

The first part: c²[3x - (p+q+r)]

For the second part: p(p·x - p·q) = p(p·x), and similarly q(q·x), r(r·x).

So the equation becomes: c²[3x - (p+q+r)] - [p(p·x) + q(q·x) + r(r·x)] = 0.

Now, since p, q, r form an orthogonal basis, any vector x can be written as x = (p·x/c²)p + (q·x/c²)q + (r·x/c²)r, so p(p·x) + q(q·x) + r(r·x) = c²x.

Substituting: c²[3x - (p+q+r)] - c²x = 0 → c²[2x

[1997]

JEE Main 1997 Mathematics question on Vector Algebra