\(0<x<6 \)
\(0<x<7 \)
\(1<x<7 \)
\(1≤x≤7\)
\(0≤x≤6\)
Given that
\(|4 - x| × |a| < |3a|\)
Here, \(|a|\) represents the magnitude of vector \(a\).
Since \(a\) is a non-zero vector, its magnitude \(|a|\) is also non-zero.
So, we can now divide both sides of the inequality by \(|a|\)
\(|4 - x| < |3|\)
Now, we have an absolute value inequality with a constant on the right side.
To solve this, we can take two cases:
Case 1:
\( 4 - x \) is positive \((4 - x > 0) |4 - x| = 4 - x\)
In this case, our inequality becomes: \(4 - x < 3\)
Now, solve for \(x\) : \(x > 4 - 3 x > 1\)
Case 2: 4 - x is negative \((4 - x < 0) |4 - x| = -(4 - x) = x - 4\)
In this case, our inequality becomes: \(x - 4 < 3\)
Now, solve for \( x: x < 3 + 4 x < 7\)
So, the valid values of x in this case are \(1 < x < 7.\)
Combining both cases, we find that the valid values of \(x \) are \(1 < x < 7.\) (_Ans)\(\)
Let \(a=i+j+2k\) and \(b=i-2j+3k\) be two vectors. Then the unit vector in the direction of \(a-b\) is