JEE Main 2021 (Online) 25th July Morning Shift | Vector Algebra Question 18 | Physics

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JEE Main 2021 (Online) 25th July Morning Shift

MCQ (Single Correct Answer)

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Match List - I with List - II

JEE Main 2021 (Online) 25th July Morning Shift Physics - Vector Algebra Question 18 English

Choose the correct answer from the options given below :

(a) $$\to$$ (iv), (b) $$\to$$ (i), (c) $$\to$$ (iii), (d) $$\to$$ (ii)

(a) $$\to$$ (iv), (b) $$\to$$ (iii), (c) $$\to$$ (i), (d) $$\to$$ (ii)

(a) $$\to$$ (iii), (b) $$\to$$ (ii), (c) $$\to$$ (iv), (d) $$\to$$ (i)

(a) $$\to$$ (i), (b) $$\to$$ (iv), (c) $$\to$$ (ii), (d) $$\to$$ (iii)

JEE Main 2021 (Online) 22th July Evening Shift

MCQ (Single Correct Answer)

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What will be the projection of vector $$\overrightarrow A = \widehat i + \widehat j + \widehat k$$ on vector $$\overrightarrow B = \widehat i + \widehat j$$ ?

$$\sqrt 2 (\widehat i + \widehat j + \widehat k)$$

$$(\widehat i + \widehat j)$$

$$\sqrt 2 (\widehat i + \widehat j)$$

$$2(\widehat i + \widehat j + \widehat k)$$

JEE Main 2021 (Online) 20th July Evening Shift

MCQ (Single Correct Answer)

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Two vectors $${\overrightarrow P }$$ and $${\overrightarrow Q }$$ have equal magnitudes. If the magnitude of $${\overrightarrow P + \overrightarrow Q }$$ is n times the magnitude of $${\overrightarrow P - \overrightarrow Q }$$, then angle between $${\overrightarrow P }$$ and $${\overrightarrow Q }$$ is :

$${\sin ^{ - 1}}\left( {{{n - 1} \over {n + 1}}} \right)$$

$${\cos ^{ - 1}}\left( {{{n - 1} \over {n + 1}}} \right)$$

$${\sin ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$$

$${\cos ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$$

JEE Main 2021 (Online) 20th July Morning Shift

MCQ (Single Correct Answer)

-1

If $$\overrightarrow A $$ and $$\overrightarrow B $$ are two vectors satisfying the relation $$\overrightarrow A $$ . $$\overrightarrow B $$ = $$\left| {\overrightarrow A \times \overrightarrow B } \right|$$. Then the value of $$\left| {\overrightarrow A - \overrightarrow B } \right|$$ will be :

$$\sqrt {{A^2} + {B^2} + \sqrt 2 AB} $$

$$\sqrt {{A^2} + {B^2}} $$

$$\sqrt {{A^2} + {B^2} - \sqrt 2 AB} $$

$$\sqrt {{A^2} + {B^2} + 2AB} $$

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