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

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(C) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, so $|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.Let $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$,

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$3$

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(C) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, so $|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.Let $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$, where $x, y, z \in \mathbb{Z}$.Given $\vec{a} \cdot \vec{b} = 1$, we have $x + y + z = 1$.Also, $\cos(	heta) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{1}{\sqrt{3}|\vec{b}|} = \frac{1}{3}$.This implies $\sqrt{3}|\vec{b}| = 3$, so $|\vec{b}| = \sqrt{3}$.Thus, $|\vec{b}|^2 = x^2 + y^2 + z^2 = 3$.We need to find integers $(x, y, z)$ such that $x + y + z = 1$ and $x^2 + y^2 + z^2 = 3$.The possible integer solutions are permutations of $(1, 1, -1)$.The permutations are $(1, 1, -1)$, $(1, -1, 1)$, and $(-1, 1, 1)$.These correspond to the vectors $\hat{i} + \hat{j} - \hat{k}$, $\hat{i} - \hat{j} + \hat{k}$, and $-\hat{i} + \hat{j} + \hat{k}$.Therefore, there are $3$ such possible vectors.

Similar Questions

The line joining the points $6 \overrightarrow{a}-4 \overrightarrow{b}+4 \overrightarrow{c}$ and $-4 \overrightarrow{c}$ and the line joining the points $-\overrightarrow{a}-2 \overrightarrow{b}-3 \overrightarrow{c}$ and $\overrightarrow{a}+2 \overrightarrow{b}-5 \overrightarrow{c}$ intersect at:

If the points whose position vectors are $2 \hat{i}+\hat{j}+\hat{k}$, $6 \hat{i}-\hat{j}+2 \hat{k}$, and $14 \hat{i}-5 \hat{j}+p \hat{k}$ are collinear, then the value of $p$ is:

The value of $x$ if $x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector is:

$I$. Two non-zero, non-collinear vectors are linearly independent.
$II$. Any three coplanar vectors are linearly dependent.
Which of the above statements is/are true?

The points with position vectors $10\,i + 3\,j$,$12\,i - 5\,j$,and $a\,i + 11\,j$ are collinear,if $a = $

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