Let vec{a}=2 hat{i}-hat{j}+hat{k} and vec{b}= | vedclass

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

Question

(D) Given: $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{j}-3 \hat{k}$.Since $\vec{a} \parallel \vec{c}$, we can write $\vec{c} = k \vec{a}$

Vedclass · Accepted Answer

$-\frac{1}{3}(5 \vec{a}+3 \vec{b})$

Vedclass · Answer

(D) Given: $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{j}-3 \hat{k}$.Since $\vec{a} \parallel \vec{c}$, we can write $\vec{c} = k \vec{a}$ for some scalar $k$.Given $\vec{b} = \vec{c} - \vec{d}$, we have $\vec{d} = \vec{c} - \vec{b} = k \vec{a} - \vec{b}$.Since $\vec{a} \perp \vec{d}$, their dot product is zero: $\vec{a} \cdot \vec{d} = 0$.$\vec{a} \cdot (k \vec{a} - \vec{b}) = 0 \Rightarrow k |\vec{a}|^2 - \vec{a} \cdot \vec{b} = 0$.Calculate $|\vec{a}|^2 = 2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6$.Calculate $\vec{a} \cdot \vec{b} = (2)(0) + (-1)(2) + (1)(-3) = 0 - 2 - 3 = -5$.Substituting these values: $k(6) - (-5) = 0 \Rightarrow 6k = -5 \Rightarrow k = -\frac{5}{6}$.Thus, $\vec{c} = -\frac{5}{6} \vec{a}$.Then $\vec{d} = \vec{c} - \vec{b} = -\frac{5}{6} \vec{a} - \vec{b}$.Finally, $\vec{c} + \vec{d} = (-\frac{5}{6} \vec{a}) + (-\frac{5}{6} \vec{a} - \vec{b}) = -\frac{10}{6} \vec{a} - \vec{b} = -\frac{5}{3} \vec{a} - \vec{b} = -\frac{1}{3}(5 \vec{a} + 3 \vec{b})$.

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{j}-3 \hat{k}$. If $\vec{b}=\vec{c}-\vec{d}$, $\vec{a}$ is parallel to $\vec{c}$, and $\vec{a}$ is perpendicular to $\vec{d}$, then $\vec{c}+\vec{d}=$

Similar Questions

If $|a| = 3$,$|b| = 4$ and the angle between $a$ and $b$ is $120^\circ$,then $|4a + 3b| = $

If $\overrightarrow{a}=\hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}-3 \hat{j}+\hat{k}$ and the orthogonal projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is $\frac{4}{3}(\hat{i}-\hat{j}-\hat{k})$,then $\lambda$ is equal to

In a triangle $ABC,$ right-angled at the vertex $A,$ if the position vectors of $A, B,$ and $C$ are respectively $3\hat{i} + \hat{j} - \hat{k},$ $-\hat{i} + 3\hat{j} + p\hat{k},$ and $5\hat{i} + q\hat{j} - 4\hat{k},$ then the point $(p, q)$ lies on a line

Let $a=\hat{i}+2 \hat{j}-2 \hat{k}$ and $b=2 \hat{i}-\hat{j}-2 \hat{k}$. If the orthogonal projection vector of $a$ on $b$ is $x$ and the orthogonal projection vector of $b$ on $a$ is $y$, then $|x-y|=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module