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

The correct option is B #160; 1 2 ( p + q + r #160;) We have #160; #160; p ×{ x q × p #160;} + q ×{ x r × q #160;} + r ×{ x p ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Orthogonal System of Vectors

Let p , q ,...

Question

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

\frac{1}{2} (\to p + \to q - 2 \to r)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2} (\to p + \to q + \to r)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{3} (\to p + \to q + \to r)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{3} (2 \to p + \to q - \to r)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{1}{2} (\to p + \to q + \to r)$
We have
$\to p \times {\to x - \to q \times \to p} + \to q \times {\to x - \to r \times \to q} + \to r \times {\to x - \to p \times \to r} = \to 0$
$\Rightarrow (\to p . \to p) (\to x - \to p) - {\to p (\to x - \to q)} \to p + (\to q . \to q) (\to x - \to r)$
$- {\to q . (\to x - \to r)} \to q + (\to r . \to r) (\to x - \to p) - {\to r . (\to x - \to p)} \to r = 0$
$\Rightarrow ({∣ ∣ \to p ∣ ∣}^{2} = {∣ ∣ \to q ∣ ∣}^{2} + {∣ ∣ \to r ∣ ∣}^{2}) \to x - ({∣ ∣ \to p ∣ ∣}^{2} \to q + {∣ ∣ \to q ∣ ∣}^{2} \to r + {∣ ∣ \to r ∣ ∣}^{2} \to p)$
$- [(\to p . \to x) \to p + (\to q . \to x) \to q + (\to r . \to x) \to r] = 0$
$\Rightarrow 3 λ^{2} \to x - λ^{2} (\to p + \to q + \to r) - {(\to p . \to x) \to p + (\to r . \to x) \to q + (\to r . \to x) \to r} = 0$ ...(1)
where $λ = ∣ ∣ \to p ∣ ∣ = ∣ ∣ \to q ∣ ∣ = ∣ ∣ \to r ∣ ∣$
Taking dot product with $\to p$ , we get
$3 λ^{2} (\to x . \to p) - λ^{2} {∣ ∣ \to p ∣ ∣}^{2} = 0 [∵ \to p . \to q = \to p . \to r = 0]$
$\Rightarrow 3 λ^{2} (\to x . \to p) - λ^{4} - λ^{2} (\to p . \to x) = 0$
$\Rightarrow \to p . \to x = \frac{λ^{2}}{2}$
Similarly , $\to q . \to x = \frac{λ^{2}}{2}$ and $\to r . \to x = \frac{λ^{2}}{2} .$
Substituting these vales in (1), we get
$3 λ^{2} \to x . λ^{2} (\to p + \to q + \to r) - \frac{λ^{2}}{2} (\to p + \to q + \to r) = \to 0$
$\Rightarrow \to x = \frac{1}{2} (\to p + \to q + \to r)$

Suggest Corrections

Similar questions

Q. Let

\to p, \to q, \to r

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

\to x

satisfies the equation

\to p \times ((\to x - \to q) \times \to p) + \to q \times ((\to x - \to r) \times \to q) + \to r \times ((\to x - \to p) \times \to r) = 0

then

\to x =

Q. Let

\to p, \to q, \to r

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

\to x

satisfies the equation

\to p [(\to x - \to q) \times \to p] + \to q \times [(\to x - \to r) \times \to q] + \to r \times [(\to x - \to p) \times \to r] = \to 0

, then

\to x

is given by

Q. For non-zero vectors

\to p, \to q

and

\to r,

let

(\to p \times \to q) \times \to r + (\to q \cdot \to r) \to q = (x^{2} + y^{2}) \to q + (14 - 4 x - 6 y) \to p

and

(\to r \cdot \to r) \to p = \to r

where

\to p

and

\to q

are non-collinear, and

x

and

y

are scalars. Then the value of

(x + y)

Q. Let

(\to p \times \to q) \times \to r + (\to q \cdot \to r) \to q = (x^{2} + y^{2}) \to q + (14 - 4 x - 6 y) \to p

and

(\to r \cdot \to r) \to p = \to r

where

\to p

and

\to q

are two non-zero non-collinear vectors, and

x

and

y

are scalars. Then the value of

(x + y)

Q. Given

\to P . \to Q = ∣ ∣ ∣ \to P \times \to Q ∣ ∣ ∣

and

\to R = \to P + \to Q

then

∣ ∣ ∣ \to R ∣ ∣ ∣

is:

Join BYJU'S Learning Program

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 $\to p, \to q, \to r$ be three mutually perpendicular vectors of the same magnitude. If a vector $\to x$ satisfies the equation $\to p \times (\to x - \to q \times \to p) + \to q \times (\to x - \to r \times \to q) + \to r \times (\to x - \to p \times \to r) = 0$ , then $\to x$ is given by