If unit vectors ^i & ^j are at right angles to each other and →p=3^i+4^j,→q=5^i,4→r=→p+→q and 2→s=→p−→q then
A
|→r+k→s|=|→r−k→s| for all real k
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B
→r is perpendicular to →s
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C
→r+→s is perpendicular to →r−→s
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D
|→r|=|→s|=|→p|=|→q|
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Solution
The correct options are A|→r+k→s|=|→r−k→s| for all real k B→r+→s is perpendicular to →r−→s C→r is perpendicular to →s Given →p=3^i+4^j,→q=5^i,4→r=→p+→q and 2→s=→p−→q 4→r=8^i+4^j⇒→r=2^i+^j 2→s=−2^i+4^j⇒→s=−^i+2^j (1). |→r+k→s|=∣∣(2−k)^i+(1+2k)^j∣∣=5k2+5 |→r−k→s|=∣∣(2+k)^i+(1−2k)^j∣∣=5k2+5 ∴|→r+k→s|=|→r−k→s| for all real k. (2) →r⋅→s=0 ∴→r is perpendicular to →s (3) (→r+→s)⋅(→r−→s)=(^i+3^j)⋅(3^i−^j)=0 ∴→r+→s is perpendicular to →r−→s (4) |→r|=|→s|=√5 |→p|=|→q|=5 Hence, options A,B and C.