Tom van Beek
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Tom van Beek
Summary
Tom van Beek is a human[1]. He was born in Maastricht[2]. He was born on December 26, 1931[3]. He passed away in Amsterdam[4]. He died on January 20, 2002[5]. He worked as an actor[6], writer[7], and radio drama actor[8].
Key Facts
- Tom van Beek was born in Maastricht[2].
- Tom van Beek died in Amsterdam[4].
- Tom van Beek was born on December 26, 1931[3].
- Tom van Beek died on January 20, 2002[5].
- Tom van Beek held citizenship in Kingdom of the Netherlands[9].
- Dutch was Tom van Beek's native language[10].
- Tom van Beek's professions included actor[6].
- Tom van Beek worked as a writer[7].
- Tom van Beek worked as a radio drama actor[8].
- Tom van Beek is recorded as male[11].
- Tom van Beek's instance of is recorded as human[12].
- Tom van Beek's Commons category is recorded as Tom van Beek[13].
- Tom van Beek's family name is recorded as van Beek[14].
- Tom van Beek's given name is recorded as Tom[15].
- Tom van Beek's described at URL is recorded as https://www.hoorspelen.eu/acteurs/beek-tom-van.html[16].
- Tom van Beek's described by source is recorded as Hoorspelen.eu[17].
- Tom van Beek's languages spoken, written or signed is recorded as Dutch[18].
- Tom van Beek's birth name is recorded as Thomas Karel van Beek[19].
- Tom van Beek's on focus list of Wikimedia project is recorded as WikiProject Dutch TV programs and their cast[20].
- Tom van Beek's copyright status as a creator is recorded as works protected by copyrights[21].
Body
Origins and Family
Born in Maastricht[2], Tom van Beek… he was born on December 26, 1931[3]. Dutch was his native language[10].
Career and Affiliations
Recorded occupations include actor[6], writer[7], and radio drama actor[8].
Death and Burial
Tom van Beek died on January 20, 2002[5]. He died in Amsterdam[4].
FAQs
Where was Tom van Beek born?
Born in Maastricht[2], Tom van Beek…
Where did Tom van Beek die?
Tom van Beek died in Amsterdam[4].
What did Tom van Beek do for work?
Tom van Beek worked as actor[6], writer[7], and radio drama actor[8].