JEE MAIN - Mathematics (2020 - 6th September Evening Slot - No. 11)
If $$\overrightarrow x $$ and $$\overrightarrow y $$ be two non-zero vectors such that
$$\left| {\overrightarrow x + \overrightarrow y } \right| = \left| {\overrightarrow x } \right|$$ and $${2\overrightarrow x + \lambda \overrightarrow y }$$ is perpendicular to $${\overrightarrow y }$$,
then the value of $$\lambda $$ is _________ .
Answer
1
Explanation
$$\left| {\overrightarrow x + \overrightarrow y } \right| = \left| {\overrightarrow x } \right|$$
Squaring both sides we get
$${\left| {\overrightarrow x } \right|^2} + 2\overrightarrow x .\overrightarrow y + {\left| {\overrightarrow y } \right|^2} = {\left| {\overrightarrow x } \right|^2}$$
$$ \Rightarrow $$ $$2\overrightarrow x .\overrightarrow y + \overrightarrow y .\overrightarrow y $$ = 0 ....(1)
Given $${2\overrightarrow x + \lambda \overrightarrow y }$$ is perpendicular to $${\overrightarrow y }$$
$$ \therefore $$ $$\left( {2\overrightarrow x + \lambda \overrightarrow y } \right).\overrightarrow y $$ = 0
$$ \Rightarrow $$ $$2\overrightarrow x .\overrightarrow y + \lambda \overrightarrow y .\overrightarrow y $$ = 0 ....(2)
Comparing (1) & (2) we get, $$\lambda $$ = 1
Squaring both sides we get
$${\left| {\overrightarrow x } \right|^2} + 2\overrightarrow x .\overrightarrow y + {\left| {\overrightarrow y } \right|^2} = {\left| {\overrightarrow x } \right|^2}$$
$$ \Rightarrow $$ $$2\overrightarrow x .\overrightarrow y + \overrightarrow y .\overrightarrow y $$ = 0 ....(1)
Given $${2\overrightarrow x + \lambda \overrightarrow y }$$ is perpendicular to $${\overrightarrow y }$$
$$ \therefore $$ $$\left( {2\overrightarrow x + \lambda \overrightarrow y } \right).\overrightarrow y $$ = 0
$$ \Rightarrow $$ $$2\overrightarrow x .\overrightarrow y + \lambda \overrightarrow y .\overrightarrow y $$ = 0 ....(2)
Comparing (1) & (2) we get, $$\lambda $$ = 1
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