Find the equation of the plane whose intercepts on the axes \(x, y, z\) are respectively \(2, 3\), and \(-4\).
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Solution and Explanation
Final answer: \[ \boxed{ 6x + 4y - 3z = 12. } \]
Top Questions on Vectors
Let $\vec{a}$ and $\vec{c}$ be unit vectors such that the angle between them is $\cos^{-1} \left( \frac{1}{4} \right)$. If $\vec{b} = 2\vec{c} + \lambda \vec{a}$. Where $\lambda > 0$ and $|\vec{b}| = 4$, then $\lambda$ is equal to:
- Let \( \mathbf{a} = e\hat{i} + 3\hat{j} + 4\hat{k}, \mathbf{b} = \hat{i} + 2\hat{j} - \hat{k}, \mathbf{c} = 3\hat{i} + \hat{j} + \lambda \hat{k} \) be the co-terminal edges of a parallelepiped whose volume is 5 units. Then the value of \( \lambda \) is:
If \( \mathbf{a} = \hat{i} + \hat{j} + \hat{k}, \, \mathbf{b} = 2\hat{i} - \hat{j} + 3\hat{k}, \, \mathbf{c} = \hat{i} - 2\hat{j} + \hat{k} \), \(\text{ then a vector of magnitude }\) \( \sqrt{22} \) \(\text{ which is parallel to }\) \( 2\mathbf{a} - \mathbf{b} + \mathbf{c} \) is:
If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}| = 3$, $|\vec{b}| = 4$ and $|\vec{a} + \vec{b}| = 1$, then the value of $|\vec{a} \times \vec{b}|$ is:
If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are three vectors such that $\vec{a} \times \vec{b} = \vec{c}$, $\vec{a} \cdot \vec{c} = 2$ and $\vec{b} \cdot \vec{c} = 1$. If $|\vec{b}| = 1$, then the value of $|\vec{a}|$ is:
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