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3-1.Vectors
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For component of a vector $A =(3 \hat{ i }+4 \hat{ j }-5 \hat{ k })$, match the following colum.
| Colum $I$ | Colum $II$ |
| $(A)$ $x-$axis | $(p)$ $5\,unit$ |
| $(B)$ Along another vector $(2 \hat{ i }+\hat{ j }+2 \hat{ k })$ | $(q)$ $4\,unit$ |
| $(C)$ Along $(6 \hat{ i }+8 \hat{ j }-10 \hat{ k })$ | $(r)$ $0$ |
| $(D)$ Along another vector $(-3 \hat{ i }-4 \hat{ j }+5 \hat{ k })$ | $(s)$ None |
A$( A \rightarrow q , B \rightarrow r , C \rightarrow s , D \rightarrow s )$
B$( A \rightarrow p , B \rightarrow r , C \rightarrow s , D \rightarrow s )$
C$( A \rightarrow r , B \rightarrow q , C \rightarrow s , D \rightarrow s )$
D$( A \rightarrow q , B \rightarrow r , C \rightarrow s , D \rightarrow p )$
Solution
(a)
$(2 \hat{ i }+\hat{ j }+4 \hat{ k })$ is perpendicular to $A$, because the dot product of these two vectors is zero.
Further $(6 \hat{ i }+4 \hat{ j }-10 \hat{ k })$ vectors is parallel to A. So, component of A along this vector is magnitude of A which is $5 \sqrt{2}$ unit. The last vector i.e. $(-3 \hat{ i }-4 \hat{ j }+5 \hat{ k })$ is anti-parallel to $A$ along this vector is negative of magnitude of A or $-5 \sqrt{2}$ unit.
$(2 \hat{ i }+\hat{ j }+4 \hat{ k })$ is perpendicular to $A$, because the dot product of these two vectors is zero.
Further $(6 \hat{ i }+4 \hat{ j }-10 \hat{ k })$ vectors is parallel to A. So, component of A along this vector is magnitude of A which is $5 \sqrt{2}$ unit. The last vector i.e. $(-3 \hat{ i }-4 \hat{ j }+5 \hat{ k })$ is anti-parallel to $A$ along this vector is negative of magnitude of A or $-5 \sqrt{2}$ unit.
Standard 11
Physics
Similar Questions
Two vectors $A$ and $B$ have equal magnitude $x$. Angle between them is $60^{\circ}$. Then, match the following two columns.
| colum $I$ | colum $II$ |
| $(A)$ $|A+B|$ | $(p)$ $\frac{\sqrt{3}}{2} x$ |
| $(B)$ $|A-B|$ | $(q)$ $x$ |
| $(C)$ $A \cdot B$ | $(r)$ $\sqrt{3} x$ |
| $(D)$ $|A \times B|$ | $(s)$ None |
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