If vec{a} = 2 hat{i} - hat{j} + hat{k},vec{b} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We are given the vectors: $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$,and $\vec{c} = \hat{i} + 3 \hat{j} -

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(A) We are given the vectors: $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$,and $\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}$.First,we calculate the vector $\lambda \vec{b} + \vec{c}$:$\lambda \vec{b} + \vec{c} = \lambda (\hat{i} + \hat{j} - 2 \hat{k}) + (\hat{i} + 3 \hat{j} - \hat{k})$$= (\lambda + 1) \hat{i} + (\lambda + 3) \hat{j} + (-2 \lambda - 1) \hat{k}$.Since $\vec{a}$ is perpendicular to $\lambda \vec{b} + \vec{c}$,their dot product must be zero:$\vec{a} \cdot (\lambda \vec{b} + \vec{c}) = 0$$(2 \hat{i} - \hat{j} + \hat{k}) \cdot ((\lambda + 1) \hat{i} + (\lambda + 3) \hat{j} + (-2 \lambda - 1) \hat{k}) = 0$$2(\lambda + 1) - 1(\lambda + 3) + 1(-2 \lambda - 1) = 0$$2 \lambda + 2 - \lambda - 3 - 2 \lambda - 1 = 0$$-\lambda - 2 = 0$$\lambda = -2$.

If $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$,and $\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}$ are given vectors. If $\vec{a}$ is perpendicular to $\lambda \vec{b} + \vec{c}$,then $\lambda = . . . . . .$.

Similar Questions

If the vectors $\bar{a}=\hat{i}-\hat{j}+2\hat{k}$,$\bar{b}=2\hat{i}+4\hat{j}+\hat{k}$ and $\bar{c}=m\hat{i}+\hat{j}+n\hat{k}$ are mutually perpendicular,then $(m, n)$ is

If $A = (k, 1, -1)$,$B = (2k, 0, 2)$,and $C = (2 + 2k, k, 1)$,and $AB \perp BC$,then the value of $k$ is:

If either vector $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0},$ then $\vec{a} \cdot \vec{b}=0 .$ But the converse need not be true. Justify your answer with an example.

Let $\hat{a}$ and $\hat{b}$ be unit vectors. If $\vec{c}$ is a vector such that the angle between $\hat{a}$ and $\vec{c}$ is $\frac{\pi}{12}$,and $\hat{b} = \vec{c} + 2(\vec{c} \times \hat{a})$,then $|6\vec{c}|^{2}$ is equal to

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}| = 2$ and $|\vec{b}| = 3$, then the maximum value of $3 |(3\vec{a} + 2\vec{b})| + 4 |(3\vec{a} - 2\vec{b})|$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module