Multiplication Of A Vector By A Scalar

$|\vec{a}+\vec{b}+\vec{c}{|}^2=|\vec{a}+\vec{b}-\vec{c}{|}^2$
$2\vec{a}\cdot \vec{b}+2\vec{b}\cdot \vec{c}+2\vec{c}\cdot \vec{a}=2\vec{a}\cdot \vec{b}-2\vec{b}\cdot \vec{c}-2\vec{c}\cdot \vec{a}$
$4\vec{a}\cdot \vec{c}=0$
$B$ is incorrect
$|\overrightarrow{a}+\lambda \overrightarrow{c}{|}^2\ge |\overrightarrow{a}{|}^2$
${\lambda }^2{c}^2\ge 0$
True $\forall \lambda \in R$ (A) is correct.

We are given that the magnitudes of the vectors and are and , and the angle between them is . We are required to find: Step 1: Using the dot product formula for the cosine of the angle between the vectors. From the given information, we know: Since , we can calculate the dot product: Step 2: Compute the cross product. Let and . The squared magnitude of is: The squared magnitude of is: Now compute the dot product : Step 3: Calculate the magnitude of the cross product. We now use the formula for the magnitude of the cross product: Substitute the values: First, expand the product : Now expand : Now subtract the two expressions: Thus, the final result is: