NEET
Practice Test 11
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The component of vector \(\dot{a}=2\hat{i}+3\hat{j}\) along the vector i + j is

  • Solution

    Component of \(\dot{a}\) along \(\dot{b}\)
    =\(=\frac{\dot{a}.\dot{b}}{\left | \dot{b} \right |}\)

The magnitude of the vector product of two vectors is \(\sqrt{3}\) times the scalar product. The angle between vectors is

  • Solution

    Here,\(\left | \dot{A}\times \dot{B} \right |=\sqrt{3}\left | \dot{A}.\dot{B} \right |\)

    \(\Rightarrow ABsin\theta =\sqrt{3}ABcos\theta \Rightarrow tan\theta =\sqrt{3}\)

    \(\Rightarrow \theta =60^{\circ}-\frac{\pi }{3}\)

If\(\left | \dot{a}+\dot{b} \right |=\left | \dot{a}-\dot{b} \right |\) then angle between \(\dot{a}\: and\: \dot{b}\) is

  • Solution

    \(\left | \dot{a}+\dot{b} \right |=\left | \dot{a}-\dot{b} \right |\Rightarrow \left | \dot{a}-\dot{b} \right |^{2}\)

    \(\Rightarrow \left | a \right |^{2}+\left | b \right |^{2}+2\dot{a}.\dot{b}=\left | a \right |^{2}+\left | b \right |^{2}-2\dot{a}.b\)

A vector \(\underset{P}{\rightarrow}_{1}\) is along the positivex-axis. If its vector product with another vector \(\underset{P}{\rightarrow}_{2}\) is zero then \(\underset{P}{\rightarrow}_{2}\) could be

Forces of 4 N and 5 N are applied at origin along X-axis and Y-axis respectively. The resultant force will be

  • Solution

    \(R=\sqrt{4^{2}+5^{2}}=\sqrt{41}N\)

    The angle θ will begiven by tanθ=5⁄4 or 0=\(tan^{-1}\left ( \frac{5}{4} \right )\)

Consider a vector F = 4 i – 3 j.Another vector that is perpendicular to F is

Let A= iA cos θ+ jA sin θ be any vector. Another vector B,which is normal to A can be expressed as

  • Solution

    The dot product should be zero.

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