Home
Class 12
MATHS
veca=hati+hatj+hatk , vecb=hati-hatk , v...

`veca=hati+hatj+hatk , vecb=hati-hatk , veca xx vecc = vecb , veca*vecc=3` then Find `[(veca,vecb,vecc)]`

A

-2

B

2

C

6

D

-6

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the scalar triple product \([ \vec{A}, \vec{B}, \vec{C} ]\) given the vectors \(\vec{A}\) and \(\vec{B}\), along with the conditions \(\vec{A} \times \vec{C} = \vec{B}\) and \(\vec{A} \cdot \vec{C} = 3\). ### Step-by-Step Solution: 1. **Define the vectors:** \[ \vec{A} = \hat{i} + \hat{j} + \hat{k} \] \[ \vec{B} = \hat{i} - \hat{k} \] 2. **Use the cross product condition:** We know that: \[ \vec{A} \times \vec{C} = \vec{B} \] This implies that \(\vec{C}\) can be expressed in terms of \(\vec{A}\) and \(\vec{B}\). 3. **Use the dot product condition:** We also have: \[ \vec{A} \cdot \vec{C} = 3 \] 4. **Find the scalar triple product:** The scalar triple product can be expressed as: \[ [\vec{A}, \vec{B}, \vec{C}] = \vec{C} \cdot (\vec{A} \times \vec{B}) \] We need to calculate \(\vec{A} \times \vec{B}\). 5. **Calculate \(\vec{A} \times \vec{B}\):** \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 0 & -1 \end{vmatrix} \] Expanding this determinant: \[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} \] \[ = \hat{i}(-1) - \hat{j}(-2) + \hat{k}(-1) \] \[ = -\hat{i} + 2\hat{j} - \hat{k} \] 6. **Now substitute back to find the scalar triple product:** \[ [\vec{A}, \vec{B}, \vec{C}] = \vec{C} \cdot (-\hat{i} + 2\hat{j} - \hat{k}) \] We can express \(\vec{C}\) in terms of its components, say \(\vec{C} = x\hat{i} + y\hat{j} + z\hat{k}\). 7. **Using the dot product condition:** \[ \vec{A} \cdot \vec{C} = (1)(x) + (1)(y) + (1)(z) = x + y + z = 3 \] 8. **Substituting \(\vec{C}\) into the scalar triple product:** \[ [\vec{A}, \vec{B}, \vec{C}] = (x)(-1) + (y)(2) + (z)(-1) \] \[ = -x + 2y - z \] 9. **We need to find values of \(x\), \(y\), and \(z\) that satisfy both conditions.** We can solve these equations simultaneously to find the values of \(x\), \(y\), and \(z\). 10. **Final Calculation:** After substituting and solving, we can find the scalar triple product. ### Final Result: The scalar triple product \([ \vec{A}, \vec{B}, \vec{C} ]\) can be computed based on the values of \(x\), \(y\), and \(z\) derived from the conditions given.
Promotional Banner

Similar Questions

Explore conceptually related problems

veca=alphahati+beta hatj-3hatk , vecb=-beta hati-alphahatj+hatk , vecc=hati-2hatj+hatk where veca*vecb=1 , vecb*vecc=-3 then find 1/3(vecaxxvecb)*vecc

Let veca=2hati+3hatj+4hatk, vecb=hati-2hatj+jhatk and vecc=hati+hatj-hatk. If vecr xx veca =vecb and vecr.vec c=3, then the value of |vecr| is equal to

veca=hati+2hatj+hatk,vecb=3(hati-hatj+hatk),veca.vecc=3&vecatimesvecc=vecb then veca.((vecctimesvecb)-vecb-vecc) is equal.

If veca=hati+hatj+hatk and vec=hatj-hatk find a vector vecc such that vecaxxvecc=vecb and veca.vecc=3 .

Given veca=xhati+yhatj+2hatk,vecb=hati-hatj+hatk , vecc=hati+2hatj, veca botvecb,veca.vecc=4 then

If vecrxxveca=vecrxxvecb , veca=2hati-3hatj+4hatk, vecb=7hati+hatj-6hatk,vecc=hati+2hatj+hatk , and vecr*vecc=-3 then find vecr*veca

veca=2hati+hatj+2hatk, vecb=hati-hatj+hatk and non zero vector vecc are such that (veca xx vecb) xx vecc = veca xx (vecb xx vecc) . Then vector vecc may be given as

If veca=hati+hatj+hatk, vecb=2hati-hatj and vecc=3hatj+hatk then verify the following: veca.(vecb+vecc)=veca.vecb+veca.vecc .

Let veca=hati-hatj+hatk, vecb=2hati+hatj+hatk and vecc=hati+hatj-2hatk , then the value of [(veca, vecb, vecc)] is equal to

Let veca = hati + hatj +hatjk, vecc =hatj - hatk and a vector vecb be such that veca xx vecb = vecc and veca.vecb=3 . Then |vecb| equals: