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Ans. (c) \hat{i} + \hat{2j}\\[4.5 bp]
Ans. (c) 7
Explanation:
Vector, V \rightarrow \space 6\hat{i} + 2\hat{j} + 3\hat{k} \\
Magnitude of the vector, V;
\\|V| = \sqrt{(6^2 + 2^2 + 3^2)} = \sqrt{(36+4+9)} = \sqrt{49} = 7
Explanation:
Two vectors can have the same magnitude.
Magnitude of the vector \hat{i} - 2\hat{j} + \hat{k} is equal to the magnitude of the vector
2\hat{i} + \hat{j} - \hat{k}.
Explanation:
A =5\hat{i} + \hat{j} - 3\hat{k} \\[2.5 bp]
B = 3\hat{i} - 4\hat{j} + 7\hat{k} \\[2.5 bp]
A . B = (5\hat{i} + \hat{j} - 3\hat{k}).(3\hat{i} - 4\hat{j} + 7\hat{k})\\[4.5 bp]
= 5 · 3 + 1 · (-4) + (-3) · 7
= 15 – 4 – 21
= -10
Explanation:
Projection of \hat{i} - \hat{j}
on the vector \hat{i} + \hat{j} \\[4.5 bp]
(\hat{i} - \hat{j}).\dfrac{(\hat{i} + \hat{j})}{|\hat{i} + \hat{j}|} = 0
Chapter No. | Chapter Name |
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Chapter 1 | Relations and Functions |
Chapter 2 | Inverse Trigonometric Functions |
Chapter 3 | Matrices |
Chapter 4 | Determinants |
Chapter 5 | Continuity and Differentiability |
Chapter 6 | Applications of Derivatives |
Chapter 7 | Integrals |
Chapter 8 | Applications of the Integrals |
Chapter 9 | Differential Equations |
Chapter 10 | Vectors |
Chapter 11 | Three - dimensional Geometry |
Chapter 12 | Linear Programming |
Chapter 13 | Probability |
Chapter Wise Important Questions for CBSE Board Class 12 Maths |
---|
Relations and Functions |
Inverse Trigonometric Functions |
Matrices |
Determinants |
Continuity and Differentiability |
Applications of Derivatives |
Integrals |
Applications of the Integrals |
Differential Equations |
Vectors |
Three - dimensional Geometry |
Linear Programming |
Probability |
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