If vectors a 1= x hat i - hat j + hat k and a 2= hat i + hat y hat j + z hat k are collinear, then a possible unit vector parallel to the vector hat hati+ hat y + z hat k is

Question

If vectors a 1= x hat i - hat j + hat k and a 2= hat i + hat y hat j + z hat k are collinear, then a possible unit vector parallel to the vector hat hati+ hat y + z hat k is (A) (1/√2)(- hat j + hat k ) (B) (1/√2)( hat i - hat j ) (C) (1/√3)( hat i + hat j - hat k ) (D) (1/√3)( hat i - hat j + hat k )

Tardigrade Admin · Accepted Answer

vec a 1 and vec a 2 are collinear so (x/1)=(-1/y)=(1/z) unit vector in direction of x hati+y hatj+z hatk=± (1/√3)( hati- hatj+ hatk)

Q. If vectors $a_{1} = x \hat{i} - \hat{j} + \hat{k}$ and $a_{2} = \hat{i} + \overset{y}{^} \hat{j} + z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $\hat{\hat{i}} + \overset{y}{^} + z \hat{k}$ is

$\frac{1}{2} (- \hat{j} + \hat{k})$

$\frac{1}{2} (\hat{i} - \hat{j})$

$\frac{1}{3} (\hat{i} + \hat{j} - \hat{k})$

$\frac{1}{3} (\hat{i} - \hat{j} + \hat{k})$

$a_{1}$ and $a_{2}$ are collinear
so $\frac{x}{1} = \frac{- 1}{y} = \frac{1}{z}$
unit vector in direction of
$x \hat{i} + y \hat{j} + z \hat{k} = \pm \frac{1}{3} (\hat{i} - \hat{j} + \hat{k})$

Q. If vectors a1​=xi^−j^​+k^ and a2​=i^+y^​j^​+zk^ are collinear, then a possible unit vector parallel to the vector i^^+y^​+zk^ is

2​1​(−j^​+k^)

2​1​(i^−j^​)

3​1​(i^+j^​−k^)

3​1​(i^−j^​+k^)